HDU 1532 Drainage Ditches 最大流模闆題

題目：http://acm.hdu.edu.cn/showproblem.php?pid=1532

題意：給定兩個數n m，分别代表圖的邊數和頂點數，然後n行形式如a b c，代表a到b之間有一條容量為c的有向邊。求1到m的最大流

思路：就是一模版題啊。。。

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

const int N = 210;
const int INF = 0x3f3f3f3f;
struct edge
{
    int to, cap, next;
}g[N*100];
int iter[N], level[N], head[N];
int n, m, cnt;
void add_edge(int v, int u, int cap)
{
    g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;
    g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++;
}
bool bfs(int s, int t)
{
    memset(level, -1, sizeof level);
    queue<int> que;
    level[s] = 0;
    que.push(s);
    while(! que.empty())
    {
        int v = que.front(); que.pop();
        for(int i = head[v]; i != -1; i =g[i].next)
        {
            int u = g[i].to;
            if(g[i].cap > 0 && level[u] < 0)
            {
                level[u] = level[v] + 1;
                que.push(u);
            }
        }
    }
    return level[t] == -1;
}
int dfs(int v, int t, int f)
{
    if(v == t) return f;
    for(int &i = iter[v]; i != -1; i = g[i].next)
    {
        int u = g[i].to;
        if(g[i].cap > 0 && level[v] < level[u])
        {
            int d = dfs(u, t, min(f, g[i].cap));
            if(d > 0)
            {
                g[i].cap -= d, g[i^1].cap += d;
                return d;
            }
        }
    }
    return 0;
}
int dinic(int s, int t)
{
    int flow = 0, f;
    while(true)
    {
        if(bfs(s, t)) return flow;
        memcpy(iter, head, sizeof head);
        while(f = dfs(s, t, INF), f > 0)
            flow += f;
    }
}
int main()
{
    int a, b, c;
    while(~ scanf("%d%d", &n, &m))
    {
        cnt = 0;
        memset(head, -1, sizeof head);
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d%d", &a, &b, &c);
            add_edge(a, b, c);
        }
        printf("%d\n", dinic(1, m));
    }
    return 0;
}