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; }
HDU 4240 Route Redundancy 一条流最大的路径
时间: 2024-10-28 23:36:09