hdu4280 Island Transport 最大流模板Dinic算法

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题意： 求从最左边的岛到最右边的岛的最大容量
这道题还学会了手动开栈
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<queue>#pragma comment(linker, "/STACK:1024000000,1024000000")//外挂开栈using namespace std;const int maxn = 100000 + 5;const int inf = 0x7f7f7f7f;int head[maxn], d[maxn];int tot;int n, m;struct Eage{    int t, w, next;}eage[maxn*2];inline void add(int x, int y, int c){    eage[tot].t = y;    eage[tot].w += c;    eage[tot].next = head[x];    head[x] = tot++;    eage[tot].t = x;    eage[tot].w += c;//注意这道题的边是双向的！    eage[tot].next = head[y];    head[y] = tot++;}bool bfs(int s, int t){    queue<int> q;    memset(d, -1, sizeof(d));    d[s] = 0;    q.push(s);    while (!q.empty())    {        int u = q.front(); q.pop();        if (u == t) return true;        for (int i = head[u]; i != -1; i = eage[i].next)        {            int v = eage[i].t;            if (d[v] == -1 && eage[i].w > 0)//i写成v调试两个多小时            {                d[v] = d[u] + 1;                q.push(v);            }        }    }    return false;}int dfs(int u, int flow, int t){    if (u == t) return flow;    int sum = 0;    for (int i = head[u]; i != -1; i = eage[i].next)    {        int v = eage[i].t;        int w = eage[i].w;        if (w > 0 && d[v] == d[u]+1) {            int tmp = dfs(v, min(flow-sum, eage[i].w), t);            eage[i].w -= tmp;            eage[i^1].w += tmp;            sum += tmp;            if (sum == flow) return sum;        }    }    if (sum == 0)d[u] = 0;    return sum;}int dinic(int s, int t){    int total = 0;    while(bfs(s, t))    {        total+=dfs(s, inf, t);    }    return  total;}int main(){    //freopen("input.txt", "r", stdin);    int t;    cin >> t;    while (t--)    {        cin >> n >> m;        int s, t, x, y;        int Min = inf, Max = -inf;        for (int i = 1; i <= n; i++)        {            scanf("%d%d", &x, &y);            if (x < Min) {                Min = x;                s = i;            }            if (x > Max)            {                Max = x;                t = i;            }        }        memset(head, -1, sizeof(head));        memset(eage, 0, sizeof(eage));        tot = 0;        for (int i = 0; i < m; i++)        {            int u, v, x;            scanf("%d%d%d", &u, &v, &x);            add(u, v, x);        }        int ans = dinic(s, t);        printf("%d\n", ans);    }    return 0;}
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