UVA - 11045 My T-shirt suits me 网络流

来源：互联网发布：韩后知乎编辑：程序博客网时间：2024/06/05 20:09

题目大意：服装的种类有6中，现给出N件服装，M个自愿者，自愿者每人都能匹配2个尺码的服装，现在问每个自愿者是否都能得到匹配到服装

解题思路：弄一个超级源点，连接者6个点，这6个点分别对应这六种服装，源点到6个点的权值都为N/6，然后在设置一个汇点，M个自愿者连接到这个汇点，权值为1，因为每个自愿者只能匹配一件衣服。最后把每个自愿者连接到所能匹配的衣服的点，权值也为1，如图，图比较难看，请见谅

#include<cstdio>#include<cstring>#include<queue>using namespace std;#define maxn 1100#define INF 0x3f3f3f3fint flow[maxn],q[maxn],head[maxn],v[maxn],next[maxn];int N,M,d[maxn],e;char str1[maxn];void add(int a,int b,int w) {v[e] = b;flow[e] = w;next[e] = head[a];head[a] = e;e++;}int bfs() {memset(d,-1,sizeof(d));int rear = 0;q[rear++] = 0;d[0] = 0;for(int i = 0; i < rear; i++)for(int j = head[q[i]]; j != -1; j = next[j]) if(d[v[j]] == -1 && flow[j]) {d[v[j]] = d[q[i]] + 1;//printf("v[j] is %d\n",v[j]);if( v[j] == 1 )return 1;q[rear++] = v[j];}return 0;}int dfs(int cur,int a) {if(cur == 1)return a;for(int i = head[cur]; i != -1; i = next[i]) if( d[v[i]] == d[cur] + 1 && flow[i]) {if(int t = dfs(v[i],a < flow[i] ? a:flow[i])) {flow[i] -= t;flow[i^1] += t;return t;}}return 0;}void init(int t) {memset(head,-1,sizeof(head));e = 0;add(0,2,t);add(2,0,0);add(0,3,t);add(3,0,0);add(0,4,t);add(4,0,0);add(0,5,t);add(5,0,0);add(0,6,t);add(6,0,0);add(0,7,t);add(7,0,0);}int EK() {int ans = 0, t;while(bfs()) {while(t = dfs(0,INF)){ans += t;}}return ans;}int main() {int test;scanf("%d",&test);while(test--) {scanf("%d%d",&N,&M);int t1;init(N / 6);for(int i = 0; i < M; i++)  {for(int j = 0; j < 2; j++) {scanf("%s",str1);if(strcmp(str1,"XXL") == 0)t1 = 2;else if(strcmp(str1,"XL") == 0)t1 = 3;else if(strcmp(str1,"L") == 0)t1 = 4;else if(strcmp(str1,"M") == 0)t1 = 5;else if(strcmp(str1,"S") == 0)t1 = 6;else if(strcmp(str1,"XS") == 0)t1 = 7;add(t1,i+8,1);add(i+8,t1,0);}add(i+8,1,1);add(1,i+8,0);}if(M - EK() == 0)   printf("YES\n");elseprintf("NO\n");}return 0;}

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