POJ-1459 Power Network 网络流

题意：给定一些散列的源点会汇点，求解网络流。

代码如下：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

const int INF = 0x3fffffff;
const int SS = 105, TT = 106;
int n, np, nc, m;

struct Edge {
    int v, c, next;    
};
Edge e[100000];
int idx, head[110];
int front, tail, que[110];
int lv[110];

void insert(int a, int b, int c) {
    e[idx].v = b, e[idx].c = c;
    e[idx].next = head[a];
    head[a] = idx++; 
}

void read_3(int &a, int &b, int &c) {
    char ch;
    while (ch = getchar(), ch != '(') ;
    scanf("%d,%d)%d", &a, &b, &c);
//    printf("%d %d %d\n", a, b, c);
}

void read_2(int &a, int &c) {
    char ch;
    while (ch = getchar(), ch != '(') ;
    scanf("%d)%d", &a, &c);
//    printf("%d %d\n", a, c);
}

bool bfs() {
    memset(lv, 0xff, sizeof (lv));
    front = tail = 0;
    lv[SS] = 0;
    que[tail++] = SS;
    while (front < tail) {
        int u = que[front++];
        for (int i = head[u]; i != -1; i = e[i].next) {
            if (!(~lv[e[i].v]) && e[i].c) {
                lv[e[i].v] = lv[u] + 1;
                if (e[i].v == TT) return true;
                que[tail++] = e[i].v;
            }
        }
    }
    return ~lv[TT];
}

int dfs(int u, int sup) {
    if (u == TT) return sup;
    int tf = 0, f;
    for (int i = head[u]; i != -1; i = e[i].next) {
        if (lv[u]+1==lv[e[i].v] && e[i].c && (f=dfs(e[i].v, min(e[i].c, sup-tf)))) {
            tf += f;
            e[i].c -= f, e[i^1].c += f;
            if (tf == sup) return sup;    
        }
    }
    if (!tf) lv[u] = -1;
    return tf;
}

int dinic() {
    int ret = 0;
    while (bfs()) {
        ret += dfs(SS, INF);
    //    printf("ret = %d\n", ret);
    }
    return ret;
}

int main() {
    int a, b, c;
    while (scanf("%d %d %d %d", &n, &np, &nc, &m) != EOF) {
        idx = 0;
        memset(head, 0xff, sizeof (head));
        for (int i = 0; i < m; ++i) { // m条边 
            read_3(a, b, c);
            insert(a, b, c);
            insert(b, a, 0);
        }
        for (int i = 0; i < np; ++i) {
            read_2(a, c);
            insert(SS, a, c);
            insert(a, SS, 0);
        }
        for (int i = 0; i < nc; ++i) {
            read_2(a, c);
            insert(a, TT, c);
            insert(TT, a, 0);
        }
        printf("%d\n", dinic());
    }
    return 0;    
}