本文主要是介绍【POJ3268】【Silver Cow Party】【反向dij】【sizeof失效】，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 15522 | Accepted: 7039 |
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
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
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
Source
sizeof() 失效
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;int N,M,X;
int g[1010][1010];
int d1[1010];
int d2[1010];
bool vis[1010];
const int INF = 0x7fffffff/100;
void dij(int g[1010][1010],int d[1010])
{rep(i,0,N) d[i] = INF;memset(vis,0,sizeof(vis));d[X] = 0;rep(i,0,N-1){int dmin = INF;int p = -1;rep(j,0,N) {if(!vis[j] && d[j] < dmin){dmin = d[j];p = j;}}vis[p] = true;rep(j,0,N) {if(!vis[j] && d[j] > d[p] + g[p][j]){d[j] = d[p] + g[p][j];}}}
}void travs()
{rep(i,0,N) rep(j,i,N) {int tmp = g[i][j];g[i][j] = g[j][i];g[j][i] = tmp;}
}
int main()
{while(scanf("%d%d%d",&N,&M,&X) != EOF){X --;rep(i,0,N) rep(j,0,N) g[i][j] = INF;rep(i,0,M) {int u,v,w;scanf("%d%d%d",&u,&v,&w);u--,v--;g[u][v] = min(g[u][v],w);}dij(g,d1);travs();dij(g,d2);int maxs = -1;for(int i=0;i<N;i++){if(maxs < d1[i] + d2[i]){maxs = d1[i] + d2[i];}}printf("%d\n",maxs);}return 0;
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