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USACO 之 Section 2.1 （已解决）

时间：2016-03-24 19:56:10      阅读：334      评论：0      收藏：0      [点我收藏+]

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The Castle：

/*
　　搜索 1A
*/

 

  1 /*
  2 ID: Jming
  3 PROG: castle
  4 LANG: C++
  5 */
  6 #include <iostream>
  7 #include <fstream>
  8 #include <sstream>
  9 #include <cstdlib>
 10 #include <cstdio>
 11 #include <cstddef>
 12 #include <iterator>
 13 #include <algorithm>
 14 #include <string>
 15 #include <locale>
 16 #include <cmath>
 17 #include <vector>
 18 #include <cstring>
 19 #include <map>
 20 #include <utility>
 21 #include <queue>
 22 #include <stack>
 23 #include <set>
 24 #include <functional>
 25 using namespace std;
 26 typedef pair<int, int> PII;
 27 typedef long long int64;
 28 const int INF = 0x3f3f3f3f;
 29 const int modPrime = 3046721;
 30 const double eps = 1e-9;
 31 const int MaxN = 55;
 32 const int MaxM = 55;
 33 
 34 int castle[MaxN][MaxM];
 35 bool visited[MaxN][MaxM];
 36 int N, M;
 37 const int direction[4] = { 1, 2, 4, 8 }; // 西 北 东 南
 38 const int step[4][2] = { {0, -1}, {-1, 0}, {0, 1}, {1, 0} };
 39 
 40 struct Node
 41 {
 42     // (x, y)代表房间坐标 
 43     // z代表拆的(x, y)房间哪面墙，z只会代表两种可能N,E
 44     int x, y, z;
 45 };
 46 
 47 bool Cmp(const Node &nd1, const Node &nd2)
 48 {
 49     if (nd1.y == nd2.y)
 50     {
 51         if (nd1.x == nd2.x)
 52         {
 53             return (nd1.z < nd2.z);
 54         }
 55         else
 56         {
 57             return (nd1.x > nd2.x);
 58         }
 59     }
 60     else
 61     {
 62         return (nd1.y < nd2.y);
 63     }
 64 }
 65 
 66 void searchRoom(int x, int y, int &roomSize)
 67 {
 68     if (!visited[x][y])
 69     {
 70         ++roomSize;
 71         visited[x][y] = true;
 72         for (int i = 0; i < 4; ++i)
 73         {
 74             if (!(castle[x][y] & direction[i]))
 75             {
 76                 int x1 = x + step[i][0];
 77                 int y1 = y + step[i][1];
 78                 if ((x1 >= 0) && (x1 < M) && (y1 >= 0) && (y1 < N))
 79                 {
 80                     //cout << x1 << "  " << y1 << endl;
 81                     searchRoom(x1, y1, roomSize);
 82                 }
 83             }
 84         }
 85     }
 86 }
 87 
 88 void removeWall()
 89 {
 90     vector<Node> vecNd;
 91     int mergeRoomSize = 0;
 92     for (int x = 0; x < M; ++x)
 93     {
 94         for (int y = 0; y < N; ++y)
 95         {
 96             for (int i = 1; i <= 2; ++i) // 只要考虑去拆一个房间的北面、东面墙就好了
 97             {
 98                 if (castle[x][y] & direction[i])
 99                 {
100                     memset(visited, false, sizeof(visited));
101                     castle[x][y] ^= direction[i];
102                     int roomSize = 0;
103                     searchRoom(x, y, roomSize);
104                     if (roomSize >= mergeRoomSize)
105                     {
106                         if (roomSize > mergeRoomSize)
107                         {
108                             vecNd.clear();
109                             mergeRoomSize = roomSize;
110                         }
111                         Node nd;
112                         nd.x = x;
113                         nd.y = y;
114                         nd.z = i;
115                         vecNd.push_back(nd);
116                     }
117                     castle[x][y] ^= direction[i];
118                 }
119             }
120         }
121     }
122     printf("%d\n", mergeRoomSize);
123     sort(vecNd.begin(), vecNd.end(), Cmp);
124     printf("%d %d ", vecNd[0].x + 1, vecNd[0].y + 1);
125     if (vecNd[0].z == 1)
126     {
127         printf("N\n");
128     }
129     else
130     {
131         printf("E\n");
132     }
133 }
134 
135 void Solve()
136 {
137     int roomSum = 0;
138     int roomMaxSize = 0;
139     for (int x = 0; x < M; ++x)
140     {
141         for (int y = 0; y < N; ++y)
142         {
143             if (!visited[x][y])
144             {
145                 //cout << endl << endl << x << "  " << y << endl;
146                 ++roomSum;
147                 int roomSize = 0;
148                 searchRoom(x, y, roomSize);
149                 roomMaxSize = max(roomMaxSize, roomSize);
150             }
151         }
152     }
153     printf("%d\n", roomSum);
154     printf("%d\n", roomMaxSize);
155 
156 }
157 
158 int main()
159 {
160 #ifdef HOME
161     freopen("in", "r", stdin);
162     //freopen("out", "w", stdout);
163 #endif
164 
165     freopen("castle.in", "r", stdin);
166     freopen("castle.out", "w", stdout);
167 
168     scanf("%d %d", &N, &M);
169     memset(visited, false, sizeof(visited));
170     for (int i = 0; i < M; ++i)
171     {
172         for (int j = 0; j < N; ++j)
173         {
174             scanf("%d", &castle[i][j]);
175         }
176     }
177     Solve();
178     removeWall();
179 
180 #ifdef HOME
181     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
182 #endif
183     return 0;
184 }

 

Ordered Fractions：

解法一：

/*
　　枚举+map
*/

 1 /*
 2 ID: Jming
 3 PROG: frac1
 4 LANG: C++
 5 */
 6 #include <iostream>
 7 #include <fstream>
 8 #include <sstream>
 9 #include <cstdlib>
10 #include <cstdio>
11 #include <cstddef>
12 #include <iterator>
13 #include <algorithm>
14 #include <string>
15 #include <locale>
16 #include <cmath>
17 #include <vector>
18 #include <cstring>
19 #include <map>
20 #include <utility>
21 #include <queue>
22 #include <stack>
23 #include <set>
24 #include <functional>
25 using namespace std;
26 typedef pair<int, int> PII;
27 typedef long long int64;
28 const int INF = 0x3f3f3f3f;
29 const int modPrime = 3046721;
30 const double eps = 1e-9;
31 const int MaxN = 55;
32 const int MaxM = 55;
33 
34 struct Fraction
35 {
36     int numerator;
37     int denominator;
38 };
39 
40 int N;
41 
42 int Gcd(int a, int b)
43 {
44     if (0 == b) return a;
45     return Gcd(b, a%b);
46 }
47 
48 // double 比较函数需要自己写，否则在USACO测试平台上，map无法滤掉相同的double键值
49 struct classcomp {
50     bool operator() (const double& lhs, const double& rhs) const
51     {
52         return (lhs < rhs);
53     }
54 };
55 
56 void Solve()
57 {
58     map<double, Fraction, classcomp> ndMap;
59     for (int i = 1; i <= N; ++i)
60     {
61         for (int j = 0; j <= i; ++j)
62         {
63             double val = static_cast<double>(j) / static_cast<double>(i);
64             if (ndMap.find(val) == ndMap.end())
65             {
66                 Fraction fc;
67                 int gcd = Gcd(i, j);
68                 fc.numerator = j/gcd;
69                 fc.denominator = i/gcd;
70                 ndMap[val] = fc;
71             }
72         }
73     }
74     for (map<double, Fraction>::const_iterator cnt_it = ndMap.begin(); cnt_it != ndMap.end(); ++cnt_it)
75     {
76 
77         cout << (*cnt_it).second.numerator << "/" << (*cnt_it).second.denominator << endl;
78     }
79 }
80 
81 int main()
82 {
83 #ifdef HOME
84     freopen("in", "r", stdin);
85     //freopen("out", "w", stdout);
86 #endif
87 
88     freopen("frac1.in", "r", stdin);
89     freopen("frac1.out", "w", stdout);
90 
91     scanf("%d", &N);
92     Solve();
93 
94 #ifdef HOME
95     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
96 #endif
97     return 0;
98 }

解法二：

/*
　　枚举+筛选+排序
*/

 

  1 /*
  2 ID: Jming
  3 PROG: frac1
  4 LANG: C++
  5 */
  6 #include <iostream>
  7 #include <fstream>
  8 #include <sstream>
  9 #include <cstdlib>
 10 #include <cstdio>
 11 #include <cstddef>
 12 #include <iterator>
 13 #include <algorithm>
 14 #include <string>
 15 #include <locale>
 16 #include <cmath>
 17 #include <vector>
 18 #include <cstring>
 19 #include <map>
 20 #include <utility>
 21 #include <queue>
 22 #include <stack>
 23 #include <set>
 24 #include <functional>
 25 using namespace std;
 26 typedef pair<int, int> PII;
 27 typedef long long int64;
 28 const int INF = 0x3f3f3f3f;
 29 const int modPrime = 3046721;
 30 const double eps = 1e-9;
 31 const int MaxN = 55;
 32 const int MaxM = 55;
 33 
 34 struct Fraction
 35 {
 36     int numerator;
 37     int denominator;
 38 };
 39 
 40 int N;
 41 // 判断a和b是否互质
 42 bool isCoprimes(int a, int b)
 43 {
 44     int r = a%b;
 45     while (r)
 46     {
 47         a = b;
 48         b = r;
 49         r = a%b;
 50     }
 51     return (1 == b);
 52 }
 53 
 54 // 分数比较大小（从小到大排序）
 55 bool Cmp(const Fraction& f1, const Fraction& f2)
 56 {
 57     return (f1.numerator*f2.denominator - f1.denominator*f2.numerator < 0);
 58 }
 59 
 60 void Solve()
 61 {
 62     vector<Fraction> vecFra;
 63     for (int i = 1; i <= N; ++i)
 64     {
 65         for (int j = 0; j <= i; ++j)
 66         {
 67             if (isCoprimes(j, i)) // 滤掉化简相等的分数（i和j的顺序不能变，否则出现除以0的情况）
 68             {
 69                 Fraction fra;
 70                 fra.numerator = j;
 71                 fra.denominator = i;
 72                 vecFra.push_back(fra);
 73             }
 74         }
 75     }
 76     sort(vecFra.begin(), vecFra.end(), Cmp);
 77     for (int i = 0; i < vecFra.size(); ++i)
 78     {
 79         printf("%d/%d\n", vecFra[i].numerator, vecFra[i].denominator);
 80     }
 81 }
 82 
 83 int main()
 84 {
 85 #ifdef HOME
 86     freopen("in", "r", stdin);
 87     //freopen("out", "w", stdout);
 88 #endif
 89 
 90     freopen("frac1.in", "r", stdin);
 91     freopen("frac1.out", "w", stdout);
 92 
 93     scanf("%d", &N);
 94     Solve();
 95 
 96 #ifdef HOME
 97     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
 98 #endif
 99     return 0;
100 }

 

Sorting a Three-Valued Sequence ：

/*
　　具体参考图Sorting a Three-Valued Sequence.png：
　　（1）先把需要交换一次的全部找出来
　　（2）再把需要交换两次的全部找出来
　　两者次数之和即为答案
*/

Sorting a Three-Valued Sequence.png

技术分享

 

 1 /*
 2 ID: Jming
 3 PROG: sort3
 4 LANG: C++
 5 */
 6 #include <iostream>
 7 #include <fstream>
 8 #include <sstream>
 9 #include <cstdlib>
10 #include <cstdio>
11 #include <cstddef>
12 #include <iterator>
13 #include <algorithm>
14 #include <string>
15 #include <locale>
16 #include <cmath>
17 #include <vector>
18 #include <cstring>
19 #include <map>
20 #include <utility>
21 #include <queue>
22 #include <stack>
23 #include <set>
24 #include <functional>
25 using namespace std;
26 typedef pair<int, int> PII;
27 typedef long long int64;
28 const int INF = 0x3f3f3f3f;
29 const int modPrime = 3046721;
30 const double eps = 1e-9;
31 const int MaxN = 1010;
32 const int MaxM = 55;
33 
34 int seq[MaxN];
35 int ascOrder[MaxN];
36 
37 int N;
38 
39 
40 void Solve()
41 {
42     int onceCnt = 0;
43     for (int i = 0; i < N; ++i)
44     {
45         if (seq[i] != ascOrder[i])
46         {
47             for (int j = i + 1; j < N; ++j)
48             {
49                 if ((seq[j] == ascOrder[i]) &&
50                     (seq[i] == ascOrder[j]))
51                 {
52                     swap(seq[i], seq[j]);
53                     ++onceCnt;
54                     break;
55                 }
56             }
57         }
58     }
59     int twiceCnt = 0;
60     for (int i = 0; i < N; ++i)
61     {
62         if (seq[i] != ascOrder[i])
63         {
64             ++twiceCnt;
65         }
66     }
67     twiceCnt = ((twiceCnt/3)<<1);
68 
69     printf("%d\n", onceCnt + twiceCnt);
70 }
71 
72 int main()
73 {
74 #ifdef HOME
75     freopen("in", "r", stdin);
76     //freopen("out", "w", stdout);
77 #endif
78 
79     freopen("sort3.in", "r", stdin);
80     freopen("sort3.out", "w", stdout);
81 
82     scanf("%d", &N);
83     for (int i = 0; i < N; ++i)
84     {
85         scanf("%d", &seq[i]);
86         ascOrder[i] = seq[i];
87     }
88     sort(ascOrder, ascOrder + N);
89     Solve();
90 
91 
92 #ifdef HOME
93     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
94 #endif
95     return 0;
96 }

 

Healthy Holsteins：

/*
　　枚举所有饲料种类能产生的组合，判断出满足条件的组合。
　　时间复杂度：N*(2^M)
　　注：
　　　　N:the number of types of vitamins
　　　　M:the number of types of feeds available
*/

 

  1 /*
  2 ID: Jming
  3 PROG: holstein
  4 LANG: C++
  5 */
  6 #include <iostream>
  7 #include <fstream>
  8 #include <sstream>
  9 #include <cstdlib>
 10 #include <cstdio>
 11 #include <cstddef>
 12 #include <iterator>
 13 #include <algorithm>
 14 #include <string>
 15 #include <locale>
 16 #include <cmath>
 17 #include <vector>
 18 #include <cstring>
 19 #include <map>
 20 #include <utility>
 21 #include <queue>
 22 #include <stack>
 23 #include <set>
 24 #include <functional>
 25 using namespace std;
 26 typedef pair<int, int> PII;
 27 typedef long long int64;
 28 const int INF = 0x3f3f3f3f;
 29 const int modPrime = 3046721;
 30 const double eps = 1e-9;
 31 const int MaxN = 30;
 32 const int MaxM = 20;
 33 
 34 int minReq[MaxN];
 35 int feeds[MaxM][MaxN];
 36 int N;
 37 int M;
 38 int ans[MaxN];
 39 bool isRight = false;
 40 
 41 bool check(const int numOftype)
 42 {
 43     for (int i = 0; i < N; ++i)
 44     {
 45         int sum = 0;
 46         for (int j = 0; j < numOftype; ++j)
 47         {
 48             sum += feeds[ans[j]][i];
 49         }
 50         if (sum < minReq[i])
 51         {
 52             return false;
 53         }
 54     }
 55     return true;
 56 }
 57 
 58 void Dfs(int num, int pos, const int numOftype)
 59 {
 60     if (num == numOftype)
 61     {
 62         isRight = check(numOftype);
 63         return;
 64     }
 65     for (int i = pos; i < M; ++i)
 66     {
 67         ans[num] = i;
 68         Dfs(num + 1, i + 1, numOftype);
 69         if (isRight)
 70         {
 71             return;
 72         }
 73     }
 74 }
 75 
 76 
 77 void Solve()
 78 {
 79     for (int i = 1; i <= M; ++i)
 80     {
 81         Dfs(0, 0, i);
 82         if (isRight)
 83         {
 84             printf("%d ", i);
 85             sort(ans, ans + i);
 86             for (int j = 0; j < i; ++j)
 87             {
 88                 printf("%d", ans[j] + 1);
 89                 if (j != (i - 1))
 90                 {
 91                     printf(" ");
 92                 }
 93             }
 94             printf("\n");
 95             return;
 96         }
 97     }
 98 }
 99 
100 int main()
101 {
102 #ifdef HOME
103     freopen("in", "r", stdin);
104     //freopen("out", "w", stdout);
105 #endif
106 
107     freopen("holstein.in", "r", stdin);
108     freopen("holstein.out", "w", stdout);
109 
110     scanf("%d", &N);
111     for (int i = 0; i < N; ++i)
112     {
113         scanf("%d", &minReq[i]);
114     }
115     scanf("%d", &M);
116     for (int i = 0; i < M; ++i)
117     {
118         for (int j = 0; j < N; ++j)
119         {
120             scanf("%d", &feeds[i][j]);
121         }
122     }
123     Solve();
124 
125 
126 #ifdef HOME
127     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
128 #endif
129     return 0;
130 }

 

Hamming Codes：

解法一：

/*
　　枚举
　　时间复杂度：O（（(2^B)-1）*N*B）
*/

 

  1 /*
  2 ID: Jming
  3 PROG: hamming
  4 LANG: C++
  5 */
  6 #include <iostream>
  7 #include <fstream>
  8 #include <sstream>
  9 #include <cstdlib>
 10 #include <cstdio>
 11 #include <cstddef>
 12 #include <iterator>
 13 #include <algorithm>
 14 #include <string>
 15 #include <locale>
 16 #include <cmath>
 17 #include <vector>
 18 #include <cstring>
 19 #include <map>
 20 #include <utility>
 21 #include <queue>
 22 #include <stack>
 23 #include <set>
 24 #include <functional>
 25 using namespace std;
 26 typedef pair<int, int> PII;
 27 typedef long long int64;
 28 const int INF = 0x3f3f3f3f;
 29 const int modPrime = 3046721;
 30 const double eps = 1e-9;
 31 const int MaxN = 30;
 32 const int MaxM = 20;
 33 
 34 int N, B, D;
 35 vector<int> ans;
 36 
 37 bool isLegal(int x, int y)
 38 {
 39     int z = x^y;
 40     int cnt = 0;
 41     while (z)
 42     {
 43         if (z & 1)
 44         {
 45             ++cnt;
 46         }
 47         z >>= 1;
 48     }
 49     return (cnt >= D);
 50 }
 51 
 52 void Solve()
 53 {
 54     ans.push_back(0);
 55     int cnt = 1;
 56     int val = 1;
 57     while (cnt != N)
 58     {
 59         bool judge = true;
 60         for (int i = 0; i < ans.size(); ++i)
 61         {
 62             if (!isLegal(val, ans[i]))
 63             {
 64                 judge = false;
 65                 break;
 66             }
 67         }
 68         if (judge)
 69         {
 70             ++cnt;
 71             ans.push_back(val);
 72         }
 73         ++val;
 74     }
 75 }
 76 
 77 void output()
 78 {
 79     for (int i = 1; i <= ans.size(); ++i)
 80     {
 81         printf("%d", ans[i - 1]);
 82         if (i != ans.size())
 83         {
 84             if (i % 10)
 85             {
 86                 printf(" ");
 87             }
 88             else
 89             {
 90                 printf("\n");
 91             }
 92         }
 93     }
 94     printf("\n");
 95 }
 96 
 97 
 98 int main()
 99 {
100 #ifdef HOME
101     freopen("in", "r", stdin);
102     //freopen("out", "w", stdout);
103 #endif
104 
105     freopen("hamming.in", "r", stdin);
106     freopen("hamming.out", "w", stdout);
107     scanf("%d %d %d", &N, &B, &D);
108     Solve();
109     output();
110 
111 #ifdef HOME
112     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
113 #endif
114     return 0;
115 }

 

解法二：

/*
　　枚举
　　预先处理出codewords间的Hamming distance，保存在数组中。
*/

 

  1 /*
  2 ID: Jming
  3 PROG: hamming
  4 LANG: C++
  5 */
  6 #include <iostream>
  7 #include <fstream>
  8 #include <sstream>
  9 #include <cstdlib>
 10 #include <cstdio>
 11 #include <cstddef>
 12 #include <iterator>
 13 #include <algorithm>
 14 #include <string>
 15 #include <locale>
 16 #include <cmath>
 17 #include <vector>
 18 #include <cstring>
 19 #include <map>
 20 #include <utility>
 21 #include <queue>
 22 #include <stack>
 23 #include <set>
 24 #include <functional>
 25 using namespace std;
 26 typedef pair<int, int> PII;
 27 typedef long long int64;
 28 const int INF = 0x3f3f3f3f;
 29 const int modPrime = 3046721;
 30 const double eps = 1e-9;
 31 const int MaxN = (1<<8) + 10;
 32 const int MaxM = 20;
 33 
 34 int N, B, D;
 35 vector<int> ans;
 36 int dis[MaxN][MaxN];
 37 
 38 int hamDis(int x, int y)
 39 {
 40     int z = x^y;
 41     int cnt = 0;
 42     while (z)
 43     {
 44         if (z & 1)
 45         {
 46             ++cnt;
 47         }
 48         z >>= 1;
 49     }
 50     return cnt;
 51 }
 52 
 53 void ini()
 54 {
 55     int maxVal = (1 << B);
 56     for (int i = 0; i < maxVal; ++i)
 57     {
 58         for (int j = i + 1; j < maxVal; ++j)
 59         {
 60             dis[i][j] = hamDis(i, j);
 61         }
 62     }
 63 }
 64 
 65 void Solve()
 66 {
 67     ans.push_back(0);
 68     int cnt = 1;
 69     int val = 1;
 70     while (cnt != N)
 71     {
 72         bool judge = true;
 73         for (int i = 0; i < ans.size(); ++i)
 74         {
 75             if (dis[ans[i]][val] < D)
 76             {
 77                 judge = false;
 78                 break;
 79             }
 80         }
 81         if (judge)
 82         {
 83             ++cnt;
 84             ans.push_back(val);
 85         }
 86         ++val;
 87     }
 88 }
 89 
 90 void output()
 91 {
 92     for (int i = 1; i <= ans.size(); ++i)
 93     {
 94         printf("%d", ans[i - 1]);
 95         if (i != ans.size())
 96         {
 97             if (i % 10)
 98             {
 99                 printf(" ");
100             }
101             else
102             {
103                 printf("\n");
104             }
105         }
106     }
107     printf("\n");
108 }
109 
110 
111 int main()
112 {
113 #ifdef HOME
114     freopen("in", "r", stdin);
115     //freopen("out", "w", stdout);
116 #endif
117 
118     freopen("hamming.in", "r", stdin);
119     freopen("hamming.out", "w", stdout);
120     scanf("%d %d %d", &N, &B, &D);
121     ini();
122     Solve();
123     output();
124 
125 #ifdef HOME
126     cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
127 #endif
128     return 0;
129 }

 

USACO 之 Section 2.1 （已解决）

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原文地址：http://www.cnblogs.com/shijianming/p/5316715.html
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