poj 1459(dinic 非递归的一个程序。。。。)

最近一直在写数论。。。。上课无聊学了一下网络流。。先刷一下这个水题吧。。。刚开始写了一个1s++效率非常低。。。只好借鉴被人的程序了。。。。。网络流第一次写。。。感觉这个算法很好的。。。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int map1[105][105], lev[105];
int que[300], iq;

bool bfs(int s, int t) {
    int u, k, i;
    memset(lev, 0, sizeof (lev));
    iq = 0, lev[s] = 1;
    que[iq++] = s;
    for (i = 0; i < iq; i++) {
        u = que[i];
        if (u == t) return 1;
        for (k = 0; k <= t; k++)
            if (map1[u][k] > 0 && !lev[k]) {
                lev[k] = lev[u] + 1, que[iq++] = k;
            }
    }
    return 0;
}

int dinic(int s, int t) {
    int a[105], cur[105], pre[105];
    int Mflow = 0, i, u, k, f, v;
    while (bfs(s, t)) {
        for (i = 0; i <= t; i++) // 初始化，回溯的时候找到cur，初始为0
            cur[i] = 0, a[i] = INT_MAX; //a里是剩余流量
        u = s;
        while (1) {
            for (v = cur[u]; v <= t; v++)
                if (map1[u][v] > 0 && lev[v] == lev[u] + 1)
                    break;
            if (v <= t) {
                cur[u] = v + 1, pre[v] = u, a[v] = map1[u][v];
                if (a[v] > a[u])
                    a[v] = a[u];
                u = v;
                if (u == t) {
                    f = a[t];
                    Mflow += f;
                    for (v = t; v != s; v = pre[v]) {
                        cur[pre[v]] = v;
                        map1[pre[v]][v] -= f, map1[v][pre[v]] += f;
                        a[v] -= f;
                        if (map1[pre[v]][v] == 0)
                            u = pre[v];
                    }
                }
            } else {
                if (u != s)
                    lev[u] = INT_MAX, u = pre[u];
                else
                    break;
            }
        }
    }
    return Mflow;
}

int main() {
    int n, np, nc, m, i, j, u, v, w;
    while (scanf("%d%d%d%d", &n, &np, &nc, &m) != -1) {
        memset(map1, 0, sizeof (map1));
        for (i = 1; i <= m; i++) {
            scanf(" (%d,%d)%d", &u, &v, &w);
            map1[u][v] = w;
        }
        for (i = 1; i <= np; i++) {
            scanf(" (%d)%d", &u, &w);
            map1[n][u] = w;
        }
        for (i = 1; i <= nc; i++) {
            scanf(" (%d)%d", &u, &w);
            map1[u][n + 1] = w;
        }
        printf("%d\n", dinic(n, n + 1));
    }
    return 0;
}