标签：width inf memory rip art rect i++ return sam
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 35984 | Accepted: 16104 |
Description
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?
Input
Output
Sample Input
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
Sample Output
10
Hint
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <algorithm> #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdio.h> #include <queue> #include <string.h> #include <vector> #include <map> #define ME(x , y) memset(x , y , sizeof(x)) #define SF(n) scanf("%d" , &n) #define rep(i , n) for(int i = 0 ; i < n ; i ++) #define INF 0x3f3f3f3f #define mod 1000000007 using namespace std; typedef long long ll ; int n , m ; int ma[1109][1109]; int dis[1109]; int vis[1109] ; int d[1109]; void Dijia(int r) { for(int i = 1 ; i <= n ; i++) { vis[i] = 0 ; dis[i] = ma[r][i]; } vis[r] = 1 ; for(int i = 1 ; i < n ; i++) { int min1 = INF ; int pos ; for(int j = 1 ; j <= n ; j++) { if(!vis[j] && dis[j] < min1) { min1 = dis[j]; pos = j ; } } vis[pos] = 1 ; for(int j = 1 ; j <= n ; j++) { if(!vis[j] && dis[j] > dis[pos] + ma[pos][j]) { dis[j] = dis[pos] + ma[pos][j]; } } } } int main() { int to ; scanf("%d%d%d" , &n , &m , &to); int u , v , w ; memset(ma , INF , sizeof(ma)); int ans = 0 ; for(int i = 1 ; i <= n ; i++) ma[i][i] = 0 ; for(int i = 1 ; i <= m ; i++) { scanf("%d%d%d" , &u , &v , &w); ma[u][v] = min(w , ma[u][v]); } Dijia(to); for(int i = 1 ; i <= n ; i++) { d[i] = dis[i]; } for(int i = 1 ; i <= n ; i++) { for(int j = i + 1 ; j <= n ; j++) { int t ; t = ma[i][j]; ma[i][j] = ma[j][i]; ma[j][i] = t ; } } Dijia(to); for(int i = 1 ; i <= n ; i++) { ans = max(ans , d[i] + dis[i]); } printf("%d\n" , ans); return 0 ; }
标签：width inf memory rip art rect i++ return sam
原文地址：https://www.cnblogs.com/nonames/p/11385747.html
