題目描述：網路流裸題

問題描述

　　一個有向圖，求1到N的最大流

輸入格式

　　第一行N M，表示點數與邊數
　　接下來M行每行s t c表示一條從s到t的容量為c的邊

輸出格式

　　一個數最大流量

樣例輸入

6 10
1 2 4
1 3 8
2 3 4
2 4 4
2 5 1
3 4 2
3 5 2
4 6 7
5 4 6
5 6 3

樣例輸出

思路

看一下大佬講的網路流吧，看完就懂了：https://blog.csdn.net/A_Comme_Amour/article/details/79356220

程式碼

// 演算法訓練：網路流裸題

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1000 + 10;
const int INF = 0x7fffffff;
int n, m, g[maxn][maxn];
int maxflow = 0;
int increase = 0;
int pre[maxn], flow[maxn];
queue<int> q;

int bfs(int s, int t)
{
    while (!q.empty())
        q.pop();
    q.push(1);
    for (int i = 1; i <= t; i++)
    {
        pre[i] = -1;
    }
    pre[s] = 0;
    flow[s] = INF;
    int ok = false;
    while (!q.empty())
    {
        int a = q.front();
        q.pop();
        if (a == t)
            break;
        for (int i = 1; i <= t; i++)
        {
            if (g[a][i] > 0 && pre[i] == -1)
            {
                pre[i] = a;
                flow[i] = min(flow[a], g[a][i]);
                q.push(i);
                if (i == t)
                {
                    ok = true;
                    break;
                }
            }
        }
        if (ok)
            break;
    }
    if (pre[t] == -1)
        return -1;
    else
        return flow[t];
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i++)
    {
        int s, t, c;
        scanf("%d%d%d", &s, &t, &c);
        g[s][t] += c;
    }
    while ((increase = bfs(1, n)) != -1)
    {
        maxflow += increase;
        int b = n;
        while (b != 1)
        {
            g[b][pre[b]] += increase;
            g[pre[b]][b] -= increase;
            b = pre[b];
        }
    }
    printf("%d", maxflow);
    return 0;
}