hdoj 4292 Food 【拆点 网络流】
2015-08-28 22:36
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FoodTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3851 Accepted Submission(s): 1289 Problem Description You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible. The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly. You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink. Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service. Input There are several test cases. For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink. The second line contains F integers, the ith number of which denotes amount of representative food. The third line contains D integers, the ith number of which denotes amount of representative drink. Following is N line, each consisting of a string of length F. e jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no. Following is N line, each consisting of a string of length D. e jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no. Please process until EOF (End Of File). Output For each test case, please print a single line with one integer, the maximum number of people to be satisfied. Sample Input 4 3 3 1 1 1 1 1 1 YYN NYY YNY YNY YNY YYN YYN NNY Sample Output 3 |
N*F矩阵的第i行第j列表示第i个人是否喜欢第j种食物,若该位置的元素为Y表示喜欢否则不喜欢。
[b]N*D矩阵的第i行第j列表示第i个人是否喜欢第j种饮料,若该位置的元素为Y表示喜欢否则不喜欢。
[/b]
每个人的需求是——选择一份食物和一份饮料(必须是他们喜欢的),问最多可以满足几个人的需求。
简单题,直接说下建图吧。
建图:超级源点source,超级汇点sink
1,source向每一种食物建边,容量为当前食物的数目;
2,每一种饮料向sink建边,容量为当前饮料的数目;
3,把每个人 i 拆成 左点i到右点i+N的一条边,容量为1,表示每个人只能选择一份食物和一份饮料。
4,每一种食物向喜欢它的人建边,食物所对应的点——>拆点后人的左点,容量为1;
5,每一个人向他喜欢的饮料建边,拆点后人的右点——>饮料所对应的点,容量为1;
最后跑一次最大流就ok了。
AC代码:
#include <cstdio> #include <cstring> #include <algorithm> #include <queue> #define MAXN 1000 #define MAXM 200000+10 #define INF 0x3f3f3f3f using namespace std; struct Edge { int from, to, cap, flow, next; }; Edge edge[MAXM]; int head[MAXN], edgenum; int dist[MAXN], cur[MAXN]; bool vis[MAXN]; int N, F, D; int source, sink;//超级源点 超级汇点 void init() { edgenum = 0; memset(head, -1, sizeof(head)); } void addEdge(int u, int v, int w) { Edge E1 = {u, v, w, 0, head[u]}; edge[edgenum] = E1; head[u] = edgenum++; Edge E2 = {v, u, 0, 0, head[v]}; edge[edgenum] = E2; head[v] = edgenum++; } void getMap() { int a; source = 0, sink = 2*N + F + D + 1; //人拆点后 编号从1到2*N for(int i = 1; i <= N; i++)//对人拆点 addEdge(i, i+N, 1);//只能选一种食物和一种饮料 //食物编号从2*N+1到2*N+F for(int i = 1; i <= F; i++) { scanf("%d", &a); addEdge(source, 2*N+i, a);//源点 到 食物建边 } //饮料编号2*N+F+1到2*N+F+D for(int i = 1; i <= D; i++) { scanf("%d", &a); addEdge(2*N+F+i, sink, a);//饮料 到 汇点建边 } char str[300]; for(int i = 1; i <= N; i++) { scanf("%s", str); for(int j = 0; j < F; j++) { if(str[j] == 'Y')//第i个人 接受 第j+1种食物 addEdge(2*N+j+1, i, 1); } } for(int i = 1; i <= N; i++) { scanf("%s", str); for(int j = 0; j < D; j++) { if(str[j] == 'Y')//第i个人 接受 第j+1种饮料 addEdge(i+N, 2*N+F+j+1, 1); } } } bool BFS(int s, int t) { queue<int> Q; memset(dist, -1, sizeof(dist)); memset(vis, false, sizeof(vis)); dist[s] = 0; vis[s] = true; Q.push(s); while(!Q.empty()) { int u = Q.front(); Q.pop(); for(int i = head[u]; i != -1; i = edge[i].next) { Edge E = edge[i]; if(!vis[E.to] && E.cap > E.flow) { dist[E.to] = dist[u] + 1; if(E.to == t) return true; vis[E.to] = true; Q.push(E.to); } } } return false; } int DFS(int x, int a, int t) { if(x == t || a == 0) return a; int flow = 0, f; for(int &i = cur[x]; i != -1; i = edge[i].next) { Edge &E = edge[i]; if(dist[E.to] == dist[x] + 1 && (f = DFS(E.to, min(a, E.cap - E.flow), t)) > 0) { edge[i].flow += f; edge[i^1].flow -= f; flow += f; a -= f; if(a == 0) break; } } return flow; } int Maxflow(int s, int t) { int flow = 0; while(BFS(s, t)) { memcpy(cur, head, sizeof(head)); flow += DFS(s, INF, t); } return flow; } int main() { while(scanf("%d%d%d", &N, &F, &D) != EOF) { init(); getMap(); printf("%d\n", Maxflow(source, sink)); } return 0; }
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