uva 753(最大流)

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题意：有n个插座，m个插头，k种不同的转换器，只能相同名字的插头和插座才能匹配，问最少剩余多少个插头无法匹配。

题解：最大流问题，建图时可以将汇点的下标位置先保存下来，最后所有型号的字符串都保存完毕，再设一个超级汇点将之前的汇点汇到一起。

#include <stdio.h>#include <string.h>#include <queue>using namespace std;const int N = 500;const int INF = 0x3f3f3f3f;int t, n, cap[N][N], m, k, cnt, maxflow, flow[N][N], p[N], a[N];char str[N][30];int ek(int en) {queue<int> q;memset(flow, 0, sizeof(flow));maxflow = 0;while (1) {memset(a, 0, sizeof(a));a[0] = INF;q.push(0);while (!q.empty()) {int u = q.front();q.pop();for (int v = 0; v <= cnt; v++) {if (!a[v] && cap[u][v] > flow[u][v]) {a[v] = a[u] < cap[u][v] - flow[u][v] ? a[u] : cap[u][v] - flow[u][v];p[v] = u;q.push(v);}}}if (a[en] == 0)break;for (int i = en; i != 0; i = p[i]) {flow[p[i]][i] += a[en];flow[i][p[i]] -= a[en];}maxflow += a[en];}}int find(char *p) {for (int i = 1; i < cnt; i++) {if (strcmp(str[i], p) == 0)return i;}strcpy(str[cnt], p);return cnt++;}int main() {scanf("%d", &t);while (t--) {scanf("%d", &n);memset(cap, 0, sizeof(cap));cnt = 1;char temp1[30], temp2[30];for (int i = 1; i <= n; i++) {scanf("%s", temp1);int a = find(temp1);cap[0][a]++;}int temp[N];scanf("%d", &m);for (int i = 0; i < m; i++) {scanf("%s%s", temp2, temp1);temp[i] = find(temp1);}scanf("%d", &k);for (int i = 0; i < k; i++) {scanf("%s%s", temp1, temp2);int x1 = find(temp1);int x2 = find(temp2);cap[x2][x1] = INF;}for (int i = 0; i < m; i++)cap[temp[i]][cnt]++;ek(cnt);printf("%d\n", m - maxflow);if (t)printf("\n");}return 0;}

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