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给定一一个n个点m条边的加权有向图, 平均值最小的回路。
二分答案,对于每个二分的mid 做一次Bellman-Fprd , 假设有k条边组成的回路。 回路上各条边的权值为 w1 , w2 ..wk ,
那么平均值小于mid意味着w1+w2+w3..+wk< k*mid 即:
(w1 - min)+(w2-mid)+...+(w2-mid)<0;
也就是说 这k条边能组成 一个负环,用 Bellman_Ford 来检查
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <string.h> 5 #include <vector> 6 #include <queue> 7 #include <cmath> 8 using namespace std; 9 const int maxn = 55; 10 int cmp(double a ,double b){ 11 if(fabs(a-b)<=0.00000001) return 0; 12 return a-b>0?1:-1; 13 } 14 struct Edge{ 15 int from,to; 16 double dist; 17 }; 18 struct BellmanFord{ 19 int n,m; 20 vector<Edge> edges; 21 vector<int> G[maxn]; 22 bool inq[maxn]; 23 double d[maxn]; 24 int p[maxn]; 25 int cnt[maxn]; 26 void inti(int n){ 27 m=0; 28 this->n = n; 29 for(int i=0; i<n; ++i ) G[i].clear(); 30 edges.clear(); 31 } 32 void AddEdge(int form, int to, double dist){ 33 edges.push_back((Edge){form,to,dist}); 34 m =edges.size(); 35 G[form].push_back(m-1); 36 } 37 bool negativeCycle(){ 38 queue<int> Q; 39 memset(inq, 0, sizeof(inq)); 40 memset(cnt, 0, sizeof(cnt)); 41 for(int i=0; i < n; ++i) { d[i] =0; inq[i] = true; Q.push(i);} 42 while(!Q.empty()){ 43 int u = Q.front(); Q.pop(); 44 inq[u] = false; 45 for(int i=0; i<G[u].size(); i++){ 46 Edge &e = edges[G[u][i]]; 47 if(cmp(d[e.to] , d[u] + e.dist)>0){ 48 d[e.to] = d[u] +e.dist; 49 p[e.to] = G[u][i]; 50 if(!inq[e.to]){ 51 Q.push(e.to); inq[e.to] = true; 52 if(++cnt[e.to]>n) return true; 53 } 54 } 55 } 56 } 57 return false; 58 } 59 }solver; 60 bool test(double x){ 61 for(int i=0; i<solver.m; i++){ 62 solver.edges[i].dist-=x; 63 } 64 bool ret = solver.negativeCycle(); 65 for(int i= 0; i<solver.m; i++) 66 solver.edges[i].dist+=x; 67 return ret; 68 } 69 int main() 70 { 71 int T; 72 scanf("%d",&T); 73 for(int kase =1; kase<=T; kase++){ 74 int n,m; 75 scanf("%d%d",&n,&m); 76 solver.inti(n); 77 int ub =0; 78 while(m--){ 79 int u,v,w; 80 scanf("%d%d%d",&u,&v,&w); u--,v--; ub= max(ub,w); 81 solver.AddEdge(u,v,w); 82 } 83 printf("Case #%d: ",kase); 84 double ans = ub; 85 if(!test(ub+1)){ 86 printf("No cycle found.\n"); 87 }else{ 88 double L =0,R= ub; 89 while(R-L>1e-3){ 90 double M = L+(R-L)/2; 91 if(test(M)){ 92 R=M; 93 }else { 94 L=M; 95 } 96 } 97 printf("%.2lf\n",L); 98 } 99 100 } 101 return 0; 102 }
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原文地址:http://www.cnblogs.com/Opaser/p/4345601.html