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这道题没有什么可说的，先统计，然后几次快排，答案就出来了
Code（整齐但不简洁的代码）
1 #include<iostream> 2 #include<cstdio> 3 #include<fstream> 4 #include<cstring> 5 #include<queue> 6 #include<algorithm> 7 using namespace std; 8 typedef bool boolean; 9 template<typename T> 10 inline void readInteger(T& u){ 11 char x; 12 while(!isdigit((x = getchar()))); 13 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘); 14 ungetc(x, stdin); 15 } 16 int m, n; 17 int k, l; 18 int d; 19 typedef class Point{ 20 public: 21 int x; 22 int y; 23 Point(const int x = 0, const int y = 0):x(x), y(y){} 24 boolean operator <(Point another) const{ 25 if(this->x != another.x) return this->x < another.x; 26 return this->y < another.y; 27 } 28 }Point; 29 inline Point readPoint(){ 30 Point a; 31 readInteger(a.x); 32 readInteger(a.y); 33 return a; 34 } 35 typedef class MyData{ 36 public: 37 int num; 38 int count; 39 MyData(const int num = 0):num(num){} 40 }MyDatas; 41 MyData *c_r; 42 MyData *c_l; 43 inline void init(){ 44 readInteger(m); 45 readInteger(n); 46 readInteger(k); 47 readInteger(l); 48 readInteger(d); 49 c_r = new MyData[(const int)(m + 1)]; 50 c_l = new MyData[(const int)(n + 1)]; 51 memset(c_r, 0, sizeof(MyData) * (m + 1)); 52 memset(c_l, 0, sizeof(MyData) * (n + 1)); 53 for(int i = 1; i <= m; i++) c_r[i].num = i; 54 for(int i = 1; i <= n; i++) c_l[i].num = i; 55 for(int i = 1; i <= d; i++){ 56 Point a = readPoint(); 57 Point b = readPoint(); 58 Point s = min(a, b); 59 if(a.x == b.x){ 60 c_l[s.y].count++; 61 }else c_r[s.x].count++; 62 } 63 } 64 boolean cmpare(const MyData &a, const MyData &b){ 65 return a.count > b.count; 66 } 67 boolean cmpare2(const MyData &a, const MyData &b){ 68 return a.num < b.num; 69 } 70 inline void solve(){ 71 sort(c_l + 1, c_l + n + 1, cmpare); 72 sort(c_r + 1, c_r + m + 1, cmpare); 73 sort(c_l + 1, c_l + l + 1, cmpare2); 74 sort(c_r + 1, c_r + k + 1, cmpare2); 75 for(int i = 1; i <= k; i++){ 76 printf("%d ", c_r[i].num); 77 } 78 putchar(‘\n‘); 79 for(int i = 1; i <= l; i++){ 80 printf("%d ", c_l[i].num); 81 } 82 } 83 int main(){ 84 freopen("seat.in", "r", stdin); 85 freopen("seat.out", "w", stdout); 86 init(); 87 solve(); 88 return 0; 89 }
这道题是有依赖的背包问题的裸体，一个主件最多有两个附件，而且附件没有属自己的附件，所以不用考虑树形dp
直接用普通的dp就行了，考虑4个状态
1）只要主件
2）只要主件和第一个附件
3）只要主件和第二个附件
4）要主件和它的两个附件
如果某个主件的第二个附件不存在呢？不管，不会影响答案，如果特判的话很耗代码。
如果dp到附件怎么办？不管，直接跳过
如果你希望运行的速度更快，可以讲n和每个v除以10，最后结果乘上10
Code（极其不简洁的代码）
1 #include<iostream> 2 #include<cstdio> 3 #include<fstream> 4 #include<cstring> 5 #include<queue> 6 #include<algorithm> 7 using namespace std; 8 ///template starts 9 typedef bool boolean; 10 template<typename T> 11 inline void readInteger(T& u){ 12 char x; 13 while(!isdigit((x = getchar()))); 14 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘); 15 ungetc(x, stdin); 16 } 17 ///template ends 18 int n; 19 int m; 20 int *v; 21 int *w; 22 int *son[2]; 23 int f[3201]; 24 int result; 25 boolean *seced; 26 inline void init(){ 27 readInteger(n); 28 readInteger(m); 29 n /= 10; 30 v = new int[(const int)(m + 1)]; 31 w = new int[(const int)(m + 1)]; 32 son[0] = new int[(const int)(m + 1)]; 33 son[1] = new int[(const int)(m + 1)]; 34 seced = new boolean[(const int)(m + 1)]; 35 memset(seced, false, sizeof(boolean) * (m + 1)); 36 memset(son[0], 0, sizeof(int) * (m + 1)); 37 memset(son[1], 0, sizeof(int) * (m + 1)); 38 for(int i = 1, a; i <= m; i++){ 39 readInteger(v[i]); 40 v[i] /= 10; 41 readInteger(w[i]); 42 readInteger(a); 43 if(a == 0) seced[i] = true; 44 if(son[0][a] == 0) son[0][a] = i; 45 else son[1][a] = i; 46 } 47 } 48 inline void solve(){ 49 seced[0] = true; 50 v[0] = w[0] = 0; 51 for(int i = 1; i <= m; i++){ 52 if(seced[i]){ 53 for(int j = n; j >= v[i]; j--){ 54 f[j] = max(f[j - v[i]] + v[i] * w[i], f[j]); 55 int s = 0; 56 int sv = 0; 57 for(int k = 0; k < 2; k++){ 58 int r = v[son[k][i]] * w[son[k][i]]; 59 s += r; 60 sv += v[son[k][i]]; 61 if(j >= v[i] + v[son[k][i]]){ 62 f[j] = max(f[j - v[i] - v[son[k][i]]] + v[i] * w[i] + r, f[j]); 63 } 64 } 65 if(j >= v[i] + sv){ 66 f[j] = max(f[j - v[i] - sv] + v[i] * w[i] + s, f[j]); 67 } 68 } 69 } 70 } 71 printf("%d", f[n] * 10); 72 } 73 int main(){ 74 freopen("budget.in", "r", stdin); 75 freopen("budget.out", "w", stdout); 76 init(); 77 solve(); 78 return 0; 79 }
这题有两个比较常见的做法，我只写其中一个，但还是都说说
1）Tarjan + 拓扑
首先用Tarjan将所有的环（强连通分量）求出来，缩成一个点，求出它的最大值和最小值
接着从起点开始，进行拓扑排序，求出答案
2）spfa
首先可以枚举一个断点，为了使答案最大，所以从起点到断点找到一个最小值，再从终点
到起点找到一个最大值，在这个最小值这个点买入，在最大值这个点卖出，用最大值减最小值
求出这个差价。
如果每个断点都去跑两次spfa肯定会超时，所以就先用两次spfa预处理，第一次spfa在原
先的图上求出从起点出发，到i这个点的最短“距”，再从反向图中，从终点开始求出到点i这个点
的最长“距”
Code（比较复杂的代码）
1 #include<iostream> 2 #include<cstdio> 3 #include<fstream> 4 #include<cstring> 5 #include<queue> 6 #include<algorithm> 7 using namespace std; 8 ///template starts 9 typedef bool boolean; 10 template<typename T> 11 inline void readInteger(T& u){ 12 char x; 13 while(!isdigit((x = getchar()))); 14 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘); 15 ungetc(x, stdin); 16 } 17 typedef class Edge{ 18 public: 19 int end; 20 int next; 21 Edge(const int end = 0, const int next = 0):end(end), next(next){} 22 }Edge; 23 typedef class MapManager{ 24 public: 25 int ce; 26 int *h; 27 Edge *edge; 28 MapManager():ce(0), h(NULL), edge(NULL){} 29 MapManager(int limit, int points):ce(0){ 30 h = new int[(const int)(points + 1)]; 31 edge = new Edge[(const int)(limit + 1)]; 32 memset(h, 0, sizeof(int) * (points + 1)); 33 } 34 inline void addEdge(int from, int end){ 35 edge[++ce] = Edge(end, h[from]); 36 h[from] = ce; 37 } 38 }MapManager; 39 ///template ends 40 int n, m; 41 int getData(boolean (*cmpare)(const int&, const int&), int a, int b){ 42 if((*cmpare)(a, b)) return a; 43 return b; 44 } 45 int *prices; 46 inline void spfa(MapManager& g, int start, int end, int *f, boolean* visited, boolean (*cmpare)(const int&, const int&)){ 47 memset(visited, false, sizeof(boolean) * (n + 1)); 48 queue<int> que; 49 que.push(start); 50 visited[start] = true; 51 while(!que.empty()){ 52 int u = que.front(); 53 que.pop(); 54 visited[u] = false; 55 for(int i = g.h[u]; i != 0; i = g.edge[i].next){ 56 int eu = g.edge[i].end; 57 int d = getData(cmpare, f[u], prices[eu]); 58 if((*cmpare)(d, f[eu])){ 59 f[eu] = d; 60 if(!visited[eu]){ 61 que.push(eu); 62 visited[eu] = true; 63 } 64 } 65 } 66 } 67 } 68 MapManager g; 69 MapManager rev_g; 70 int *maxdis; 71 int *mindis; 72 boolean *visited; 73 boolean _min(const int &a, const int &b){ return a < b; } 74 boolean _max(const int &a, const int &b){ return a > b; } 75 inline void init(){ 76 readInteger(n); 77 readInteger(m); 78 g = MapManager(m * 2, n); 79 rev_g = MapManager(m * 2, n); 80 prices = new int[(const int)(n + 1)]; 81 maxdis = new int[(const int)(n + 1)]; 82 mindis = new int[(const int)(n + 1)]; 83 visited = new boolean[(const int)(n + 1)]; 84 memset(maxdis, 0, sizeof(int) * (n + 1)); 85 memset(mindis, 0x7f, sizeof(int) * (n + 1)); 86 for(int i = 1; i <= n; i++) 87 readInteger(prices[i]); 88 for(int i = 1, a, b, c; i <= m; i++){ 89 readInteger(a); 90 readInteger(b); 91 readInteger(c); 92 g.addEdge(a, b); 93 rev_g.addEdge(b, a); 94 if(c == 2){ 95 g.addEdge(b, a); 96 rev_g.addEdge(a, b); 97 } 98 } 99 } 100 inline void solve(){ 101 spfa(g, 1, n, mindis, visited, _min); 102 spfa(rev_g, n, 1, maxdis, visited, _max); 103 int result = 0; 104 for(int i = 2; i < n; i++){ 105 result = max(result, maxdis[i] - mindis[i]); 106 } 107 printf("%d", result); 108 } 109 int main(){ 110 freopen("trade.in", "r", stdin); 111 freopen("trade.out", "w", stdout); 112 init(); 113 solve(); 114 return 0; 115 }
（ps：为了防止写两遍spfa，所以一次spfa写得比较复杂）
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原文地址：http://www.cnblogs.com/yyf0309/p/5903436.html
