HDU 4240 Route Redundancy 一条流最大的路径

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题目来源：HDU 4240 Route Redundancy

题意：求最大流与一条流最大的路径的比值前者最大流求出后者是每一条路的最小值再取大

思路：我用的是dinic 可以在DFS的时候在传递一个参数表示当前增广路可以通过最大的流量然后当x==t 到达汇点时在取他们的最大值

#include <cstdio>#include <queue>#include <vector>#include <cstring>#include <algorithm>using namespace std;const int maxn = 610;const int INF = 999999999;int ans;struct Edge{    int from, to, cap, flow;    Edge(){}    Edge(int from, int to, int cap, int flow) : from(from), to(to), cap(cap), flow(flow){}};struct Point{    double d, x, y, r;    Point(){}    Point(double d, double x, double y, double r) : d(d), x(x), y(y), r(r){}}a[maxn];int n, m, s, t;vector <Edge> edges;vector <int> G[maxn];bool vis[maxn];int d[maxn];int cur[maxn];void AddEdge(int from, int to, int cap){    edges.push_back(Edge(from, to, cap, 0));    edges.push_back(Edge(to, from, 0, 0));    m = edges.size();    G[from].push_back(m-2);    G[to].push_back(m-1); }bool BFS(){    memset(vis, 0, sizeof(vis));    queue <int> Q;    Q.push(s);    d[s] = 0;    vis[s] = 1;    while(!Q.empty())    {        int x = Q.front(); Q.pop();        for(int i = 0; i < G[x].size(); i++)        {            Edge& e = edges[G[x][i]];            if(!vis[e.to] && e.cap > e.flow)            {                vis[e.to] = 1;                d[e.to] = d[x] + 1;                Q.push(e.to);            }        }    }    return vis[t];}int DFS(int x, int a, int v){if(x == t){ans = max(ans, v);return a;}    if(a == 0)    {return 0;    }    int flow = 0, f;    for(int& i = cur[x]; i < G[x].size(); i++)    {        Edge& e = edges[G[x][i]];        if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap-e.flow), min(v, e.cap))) > 0)        {            e.flow += f;            edges[G[x][i]^1].flow -= f;            flow += f;            a -= f;            if(a == 0)                break;        }    }    return flow;}int Maxflow(){    int flow = 0;    while(BFS())    {        memset(cur, 0, sizeof(cur));        flow += DFS(s, INF, INF);    }    return flow;}int main(){    int T;    scanf("%d", &T);    while(T--)    {        int cas, k;        scanf("%d %d %d %d %d", &cas, &n, &k, &s, &t);        edges.clear();        ans = 0;        for(int i = 0; i < n; i++)            G[i].clear();        for(int i = 0; i < k; i++)        {            int u, v, w;            scanf("%d %d %d", &u, &v, &w);            AddEdge(u, v, w);        }        int flow = Maxflow();        printf("%d %.3f\n", cas, (double)flow/ans);    }    return 0;}

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