USACO 2007 “March Gold” Ranking the Cows

时间：2019-06-11 19:33:36 阅读：128 评论：0 收藏：0 [点我收藏+]

标签：poj freopen NPU ring 因此 typename unsigned ons closed

题目链接：https://www.luogu.org/problemnew/show/P2881

题目链接：https://vjudge.net/problem/POJ-3275

题目大意

　　给定标号为 1~N 这 N 个数，在给定 M 组大小关系，求还需要知道多少组大小关系才可以给这组数排序？

分析1(Floyd + bitset)

　　总共需要知道 n * (n - 1) / 2 条边，因此只要求一下现在已经有了多少条边，再减一下即可。由于大小关系有传递性，因此计数之前需要求传递闭包。

　　直接上 floyd($O(n^3)?$) 会超时，需要用bitset或手动压位，可以在$O(\frac{n^3}{w})?$求出传递闭包，其中w表示字长，为64或32。

代码如下

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
  9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 12  
 13 #define pr(x) cout << #x << " = " << x << "  "
 14 #define prln(x) cout << #x << " = " << x << endl
 15  
 16 #define LOWBIT(x) ((x)&(-x))
 17  
 18 #define ALL(x) x.begin(),x.end()
 19 #define INS(x) inserter(x,x.begin())
 20  
 21 #define ms0(a) memset(a,0,sizeof(a))
 22 #define msI(a) memset(a,inf,sizeof(a))
 23 #define msM(a) memset(a,-1,sizeof(a))
 24 
 25 #define MP make_pair
 26 #define PB push_back
 27 #define ft first
 28 #define sd second
 29  
 30 template<typename T1, typename T2>
 31 istream &operator>>(istream &in, pair<T1, T2> &p) {
 32     in >> p.first >> p.second;
 33     return in;
 34 }
 35  
 36 template<typename T>
 37 istream &operator>>(istream &in, vector<T> &v) {
 38     for (auto &x: v)
 39         in >> x;
 40     return in;
 41 }
 42  
 43 template<typename T1, typename T2>
 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 45     out << "[" << p.first << ", " << p.second << "]" << "\n";
 46     return out;
 47 }
 48 
 49 inline int gc(){
 50     static const int BUF = 1e7;
 51     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 52     
 53     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 54     return *bg++;
 55 } 
 56 
 57 inline int ri(){
 58     int x = 0, f = 1, c = gc();
 59     for(; c<48||c>57; f = c==‘-‘?-1:f, c=gc());
 60     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 61     return x*f;
 62 }
 63  
 64 typedef long long LL;
 65 typedef unsigned long long uLL;
 66 typedef pair< double, double > PDD;
 67 typedef pair< int, int > PII;
 68 typedef pair< int, PII > PIPII;
 69 typedef pair< string, int > PSI;
 70 typedef pair< int, PSI > PIPSI;
 71 typedef set< int > SI;
 72 typedef vector< int > VI;
 73 typedef vector< VI > VVI;
 74 typedef vector< PII > VPII;
 75 typedef map< int, int > MII;
 76 typedef map< int, PII > MIPII;
 77 typedef map< string, int > MSI;
 78 typedef multimap< int, int > MMII;
 79 //typedef unordered_map< int, int > uMII;
 80 typedef pair< LL, LL > PLL;
 81 typedef vector< LL > VL;
 82 typedef vector< VL > VVL;
 83 typedef priority_queue< int > PQIMax;
 84 typedef priority_queue< int, VI, greater< int > > PQIMin;
 85 const double EPS = 1e-10;
 86 const LL inf = 0x7fffffff;
 87 const LL infLL = 0x7fffffffffffffffLL;
 88 const LL mod = 1e9 + 7;
 89 const int maxN = 1e3 + 7;
 90 const LL ONE = 1;
 91 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 92 const LL oddBits = 0x5555555555555555;
 93 
 94 int N, M, cnt;
 95 bitset< maxN > m[maxN];
 96 
 97 int main(){
 98     //freopen("MyOutput.txt","w",stdout);
 99     //freopen("input.txt","r",stdin);
100     INIT();
101     N = ri();
102     M = ri();
103     For(i, 1, N) m[i][i] = 1;
104     Rep(i, M) {
105         int x, y;
106         x = ri();
107         y = ri();
108         m[x][y] = 1;
109     }
110     
111     For(i, 1, N) For(j, 1, N) if(m[j][i]) m[j] |= m[i];
112     For(i, 1, N) cnt += m[i].count();
113     cnt -= N; // 减去 i->i 的，有 N 条 
114     
115     printf("%d\n", N * (N - 1) / 2 - cnt);
116     return 0;
117 }

View Code

分析2(dfs+ bitset)

　　考虑到通过压位过的邻接矩阵求传递闭包时做了很多多余的操作，我们可以用邻接链表来求传递闭包，然后用邻接矩阵计数。复杂度可以降到$O(\frac{n * (n + m)}{w})?$。

代码如下

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3  
  4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
  5 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  6 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
  9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 12  
 13 #define pr(x) cout << #x << " = " << x << "  "
 14 #define prln(x) cout << #x << " = " << x << endl
 15  
 16 #define LOWBIT(x) ((x)&(-x))
 17  
 18 #define ALL(x) x.begin(),x.end()
 19 #define INS(x) inserter(x,x.begin())
 20  
 21 #define ms0(a) memset(a,0,sizeof(a))
 22 #define msI(a) memset(a,inf,sizeof(a))
 23 #define msM(a) memset(a,-1,sizeof(a))
 24 
 25 #define MP make_pair
 26 #define PB push_back
 27 #define ft first
 28 #define sd second
 29  
 30 template<typename T1, typename T2>
 31 istream &operator>>(istream &in, pair<T1, T2> &p) {
 32     in >> p.first >> p.second;
 33     return in;
 34 }
 35  
 36 template<typename T>
 37 istream &operator>>(istream &in, vector<T> &v) {
 38     for (auto &x: v)
 39         in >> x;
 40     return in;
 41 }
 42  
 43 template<typename T1, typename T2>
 44 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 45     out << "[" << p.first << ", " << p.second << "]" << "\n";
 46     return out;
 47 }
 48 
 49 inline int gc(){
 50     static const int BUF = 1e7;
 51     static char buf[BUF], *bg = buf + BUF, *ed = bg;
 52     
 53     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
 54     return *bg++;
 55 } 
 56 
 57 inline int ri(){
 58     int x = 0, f = 1, c = gc();
 59     for(; c<48||c>57; f = c==‘-‘?-1:f, c=gc());
 60     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
 61     return x*f;
 62 }
 63  
 64 typedef long long LL;
 65 typedef unsigned long long uLL;
 66 typedef pair< double, double > PDD;
 67 typedef pair< int, int > PII;
 68 typedef pair< int, PII > PIPII;
 69 typedef pair< string, int > PSI;
 70 typedef pair< int, PSI > PIPSI;
 71 typedef set< int > SI;
 72 typedef vector< int > VI;
 73 typedef vector< VI > VVI;
 74 typedef vector< PII > VPII;
 75 typedef map< int, int > MII;
 76 typedef map< int, PII > MIPII;
 77 typedef map< string, int > MSI;
 78 typedef multimap< int, int > MMII;
 79 //typedef unordered_map< int, int > uMII;
 80 typedef pair< LL, LL > PLL;
 81 typedef vector< LL > VL;
 82 typedef vector< VL > VVL;
 83 typedef priority_queue< int > PQIMax;
 84 typedef priority_queue< int, VI, greater< int > > PQIMin;
 85 const double EPS = 1e-10;
 86 const LL inf = 0x7fffffff;
 87 const LL infLL = 0x7fffffffffffffffLL;
 88 const LL mod = 1e9 + 7;
 89 const int maxN = 1e3 + 7;
 90 const LL ONE = 1;
 91 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 92 const LL oddBits = 0x5555555555555555;
 93 
 94 struct Edge{
 95     int from, to;
 96 };
 97 
 98 struct Vertex{
 99     VI edges;
100 };
101 
102 int N, M, cnt;
103 bitset< maxN > m[maxN], vis;
104 Edge e[maxN << 4];
105 int elen;
106 Vertex v[maxN];
107 
108 // 找到 x 号节点所能到达的所有节点 
109 void dfs(int x) {
110     vis[x] = 1;
111     foreach(i, v[x].edges) { // 结果取决于 x 的孩子节点所能到达的节点，此处相当于 m[x][y] = 1 
112         int y = e[*i].to;
113         if(!vis[y]) dfs(y);
114         m[x] |= m[y];
115     }
116 }
117 
118 int main(){
119     //freopen("MyOutput.txt","w",stdout);
120     //freopen("input.txt","r",stdin);
121     INIT();
122     N = ri();
123     M = ri();
124     For(i, 1, N) m[i][i] = 1;
125     Rep(i, M) {
126         int x, y;
127         x = ri();
128         y = ri();
129         e[++elen].from = x;
130         e[elen].to = y;
131         v[x].edges.PB(elen);
132     }
133     
134     For(i, 1, N) if(!vis[i]) dfs(i);
135     For(i, 1, N) cnt += m[i].count();
136     cnt -= N; // 减去 i->i 的，有 N 条 
137     
138     printf("%d\n", N * (N - 1) / 2 - cnt);
139     return 0;
140 }

View Code

USACO 2007 “March Gold” Ranking the Cows

标签：poj freopen NPU ring 因此 typename unsigned ons closed

原文地址：https://www.cnblogs.com/zaq19970105/p/11005295.html

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