标签：out memory original unit party pac direct ber proc
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 23426 | Accepted: 10691 |
One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.
Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow‘s return route might be different from her original route to the party since roads are one-way.
Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?
4 8 2 1 2 4 1 3 2 1 4 7 2 1 1 2 3 5 3 1 2 3 4 4 4 2 3
10
1 //2017-08-08 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #include <algorithm> 6 #include <vector> 7 #include <queue> 8 9 using namespace std; 10 11 const int N = 1010; 12 const int INF = 0x3f3f3f3f; 13 struct Edge{ 14 int v, w; 15 Edge(int _v = 0, int _w = 0):v(_v), w(_w){} 16 }; 17 vector<Edge> E[2][N]; 18 bool vis[N]; 19 int dis[2][N], cnt[N], n, m, x; 20 21 bool spfa(int s, int n, int time){ 22 memset(vis, false, sizeof(vis)); 23 memset(dis[time], INF, sizeof(dis)); 24 memset(cnt, 0, sizeof(cnt)); 25 vis[s] = true; 26 dis[time][s] = 0; 27 cnt[s] = 1; 28 queue<int> q; 29 q.push(s); 30 while(!q.empty()){ 31 int u = q.front(); 32 q.pop(); 33 vis[u] = false; 34 for(int i = 0; i < E[time][u].size(); i++){ 35 int v = E[time][u][i].v; 36 int w = E[time][u][i].w; 37 if(dis[time][v] > dis[time][u] + w){ 38 dis[time][v] = dis[time][u] + w; 39 if(!vis[v]){ 40 vis[v] = true; 41 q.push(v); 42 if(++cnt[v] > n)return false; 43 } 44 } 45 } 46 } 47 return true; 48 } 49 50 int main() 51 { 52 while(scanf("%d%d%d", &n, &m, &x)!=EOF){ 53 int a, b, c; 54 while(m--){ 55 scanf("%d%d%d", &a, &b, &c); 56 E[0][a].push_back(Edge(b, c)); 57 E[1][b].push_back(Edge(a, c)); 58 } 59 spfa(x, n, 0); 60 spfa(x, n, 1); 61 int ans = 0; 62 for(int i = 1; i <= n; i++) 63 ans = max(ans, dis[0][i]+dis[1][i]); 64 printf("%d\n", ans); 65 } 66 67 return 0; 68 }
标签：out memory original unit party pac direct ber proc
原文地址：http://www.cnblogs.com/Penn000/p/7305049.html
