uva11045

来源：互联网发布：从笔记本cpu编号知型号编辑：程序博客网时间：2024/06/05 06:51

题意:

衣服共有六种型号,现给出套数,志愿者人数,以及每个志愿者合适的型号,问能否让所有志愿者都有衣服

思路:

由于衣服共六种型号,又有中数量,可以知道每种型号几件,这是建一个起点0指向每个型号,cap为n/6,然后志愿者可穿型号流向志愿者,流量为1,然后每个志愿者流向终点,流量也为1,因为每个人最多穿一件

代码:

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<queue>#include<string>using namespace std;const int maxn = 40;const int INF = 0x3c3c3c3c;int m, n;char ch[6][4] = {"XXL", "XL", "L", "M", "S", "XS"};int change(char* s) {for(int i=0; i<=5; i++)if(strcmp(s, ch[i]) == 0) return i+1;}int p[maxn];int a[maxn];int flow[maxn][maxn];int map[maxn][maxn];void ek2() {memset(flow, 0, sizeof(flow));int ans = 0;queue<int> q;while(1) {q.push(0);memset(a, 0, sizeof(a));a[0] = INF;memset(p, 0, sizeof(p));while(!q.empty()) {int u = q.front();q.pop();for(int i=0; i<=m+7; i++) {if(!a[i] && map[u][i]>flow[u][i]) {a[i] = min(a[u], map[u][i]-flow[u][i]);p[i] = u;q.push(i);}}}if(a[m+7]==0) break;for(int i=m+7; i; i=p[i]) {flow[p[i]][i] += a[m+7];flow[i][p[i]] -= a[m+7];}ans += a[m+7];}if(ans == m) printf("YES\n");else printf("NO\n");}int main() {int kase;scanf("%d", &kase);char str[4];while(kase--) {memset(map, 0, sizeof(map));int num = 0;scanf("%d%d", &n, &m);for(int i=1; i<=6; i++) {map[0][i] = n/6;}for(int i=1; i<=m; i++){scanf("%s", str);int k = change(str);map[k][6+i] = 1;scanf("%s", str);k = change(str);map[k][6+i] = 1;}for(int i=7; i<=m+6; i++) {map[i][m+7] = 1;}ek2();}return 0;}

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