D. Going in Cycle!!
Time Limit: 3000ms
Memory Limit: 131072KB
64-bit integer IO format: %lld Java class name: MainYou are given a weighted directed graph with n vertices and m edges. Each cycle in the graph has a weight, which equals to sum of its edges. There are so many cycles in the graph with different weights. In this problem we want to find a cycle with the minimum mean.
Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with two numbers n and m. m lines follow, each has three positive number a, b, c which means there is an edge from vertex a to b with weight of c.
Output
For each test case output one line containing Case #x: followed by a number that is the lowest mean cycle in graph with 2 digits after decimal place, if there is a cycle. Otherwise print No cycle found..
- n ≤ 50
- a, b ≤ n
- c ≤ 10000000
Sample Input
2
2 1
1 2 1
2 2
1 2 2
2 1 3
Output for Sample Input
Case #1: No cycle found.
Case #2: 2.50
解题：二分+spfa负权回路判定
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #include <queue> 10 #define LL long long 11 #define INF 0x3f3f3ff 12 using namespace std; 13 const int maxn = 60; 14 const double exps = 1e-3; 15 struct arc { 16 int to; 17 double w; 18 }; 19 20 vector<arc>g[maxn]; 21 int cnt[maxn],n,m; 22 double d[maxn]; 23 bool vis[maxn]; 24 bool spfa() { 25 int i,j,v,u; 26 queue<int>q; 27 for(i = 1; i <= n; i++) { 28 q.push(i); 29 d[i] = INF; 30 vis[i] = true; 31 cnt[i] = 1; 32 } 33 d[1] = 0; 34 while(!q.empty()) { 35 u = q.front(); 36 q.pop(); 37 vis[u] = false; 38 for(i = 0; i < g[u].size(); i++) { 39 v = g[u][i].to; 40 if(d[v] > d[u]+g[u][i].w) { 41 d[v] = d[u]+g[u][i].w; 42 if(!vis[v]) { 43 vis[v] = true; 44 cnt[v]++; 45 q.push(v); 46 if(cnt[u] > n) return true; 47 48 } 49 } 50 } 51 } 52 return false; 53 } 54 void modify(double val) { 55 for(int i = 1; i <= n; i++) { 56 for(int j = 0; j < g[i].size(); j++) { 57 g[i][j].w += val; 58 } 59 } 60 } 61 bool test(double val) { 62 modify(-val); 63 bool it = spfa(); 64 modify(val); 65 return it; 66 } 67 int main() { 68 int t,i,j,u,v,k = 1; 69 double lt,rt,mid,w; 70 scanf("%d",&t); 71 while(t--) { 72 scanf("%d%d",&n,&m); 73 for(i = 0; i <= n; i++) 74 g[i].clear(); 75 lt = INF; 76 rt = 0; 77 for(i = 0; i < m; i++) { 78 scanf("%d%d%lf",&u,&v,&w); 79 g[u].push_back((arc) {v,w}); 80 if(lt > w) lt = w; 81 if(w > rt) rt = w; 82 } 83 printf("Case #%d: ",k++); 84 if(test(rt+1.0)) { 85 while(rt-lt > exps) { 86 mid = lt+(rt-lt)/2; 87 if(test(mid)) rt = mid; 88 else lt = mid; 89 } 90 printf("%.2f ",rt); 91 } else printf("No cycle found. "); 92 } 93 return 0; 94 }