The maximum flow algorithm dinic

Because a wrong min and transferred half an hour - "a tragic history of OIer"

Thinking

Luo Gu P3376 Network maximum flow

Title Description

If that, given a network diagram, and the source and sink, the network obtains its maximum flow.

Input and output formats

Input formats:

The first row contains four positive integers N, M, S, T, respectively, the number of points, there are the number of edges, number source, sink point number.

Next M lines contains three positive integers ui, vi, wi, represents the i ui article there from edge to reach VI, the right side of wi (i.e., the side of the maximum flow rate wi)

Output formats:

Line, contains a positive integer, is the maximum flow of the network.

Sample input and output

Input Sample # 1:

　　4 5 4 3
　　4 2 30
　　4 3 20
　　2 3 20
　　2 1 30
　　1 3 40

Output Sample # 1:

　　50

Explanation

Time limit:  1000ms, 128M

Data Scale: 

For 30% of the data: N <= 10, M <= 25

For 70% of the data: N <= 200, M <= 1000

To 100% of the data: N <= 10000, M <= 100000

Code

#include<cstdio>
#include<cstring>
using namespace std;

const int inf = 0x3f3f3f;

int n, m, s, t, tot = -1, ans = 0;
int st[200001], dep[10001];
int min(int a, int b) { return a < b ? a : b; }
struct node
{
    int to, val, last;
}c[200001];
void add(int u, int v, int w)
{
    c[++tot].to = v;
    c[tot].val = w;
    c[tot].last = st[u];
    st[u] = tot;
}
bool bfs()
{
    int q[10001], l = 0, r = 1;
    memset(dep, 0, sizeof dep);
    dep[s] = 1;
    q[r] = s;
    while (l != r)
    {
        int head = q[++l];
        for (int i = st[head]; i != -1; i = c[i].last)
        {
            if (c[i].val > 0 && !dep[c[i].to])
            {
                dep[c[i].to] = dep[head] + 1;
                q[++r] = c[i].to;
            }
        }
    }
    if (!dep[t])
        return false;
    return true;
}
int dfs(int u, int k)
{
    if (u == t) return k;
    for (int i = st[u]; i != -1; i = c[i].last)
    {
        if ((dep[c[i].to] == dep[u] + 1) && c[i].val)
        {
            int p = dfs(c[i].to, min(c[i].val, k));
            if (p > 0)
            {
                c[i].val -= p;
                c[i ^ 1].val += p;
                return p;
            }
        }    
    }
    return 0;
}
void dinic()
{
    while (bfs())
        while (int p = dfs(s, inf))
            ans += p;
    return;
}
int main()
{
    memset(st, -1, sizeof st);
    scanf("%d%d%d%d", &n, &m, &s, &t);
    for (int i = 0; i < m; i++)
    {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        add(u, v, w);
        add(v, u, 0);
    }
    dinic();
    printf("%d\n", ans);
    return 0;
}