HDU 4280 Island Transport 网络流

来源：互联网发布：java模块化开发思路编辑：程序博客网时间：2024/05/16 17:44

有N个岛屿，M条路线，每条路都连接两个岛屿，并且每条路都有一个最大承载人数，现在想知道从最西边的岛到最东面的岛最多能有多少人过去

可以看出是最大流的问题，而且源点汇点也都给出了，可以用dinic解决

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <queue>#define rep(i, j, k) for(int i = j; i <= k; i++)#define maxn 100009#define inf 0x7fffffffusing namespace std;int s, t;int n, m, to[maxn * 3], head[maxn], done[maxn], d[maxn], tot, flow[3 * maxn], Next[3 * maxn];int cur[maxn];void add (int u, int v, int w){    to[++tot] = v;    Next[tot] = head[u];    flow[tot] = w;    head[u] = tot;}bool bfs (){    memset (done, 0, sizeof (done));    done[s] = 1;    d[s] = 1;    queue <int> q;    q.push (s);    while (!q.empty ())    {        int now = q.front ();        q.pop ();        for (int i = head[now]; i; i = Next[i])            if (!done[to[i]] && flow[i] > 0)            {                d[to[i]] = d[now] + 1;                done[to[i]] = 1;                q.push (to[i]);            }    }    return done[t];}int dfs (int x, int a){    if (x == t || a == 0)        return a;    int ret = 0, f;    for (int &i = cur[x]; i; i = Next[i])        if (d[to[i]] == d[x] + 1 && flow[i] > 0 && (f = dfs (to[i], min (flow[i], a))) > 0)        {            flow[i] -= f;            flow[i ^ 1] += f;            a -= f;            ret += f;            if (!a)                break;        }    return ret;}int MaxFlow (){    int ret = 0;    while (bfs ())    {        rep (i, 1, n)            cur[i] = head[i];        ret += dfs (s, inf);    }    return ret;}int main (){    int ti;    cin >> ti;    while (ti--)    {        int u, v, w, Min = inf, Max = -inf;        memset (head, 0, sizeof (head));        tot = 1;        cin >> n >> m;        rep (i, 1, n)        {            scanf ("%d%d", &u, &v);            if (u < Min)                Min = u, s = i;            if (u > Max)                Max = u, t = i;        }        rep (i, 1, m)            scanf ("%d%d%d", &u, &v, &w), add (u, v, w), add (v, u, w);        cout << MaxFlow () << endl;    }    return 0;}

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