POJ 3268：Silver Cow Party 求单点的来回最短路径

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15989		Accepted: 7303

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 irequires T_i (1 ≤ T_i ≤ 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

Line 1: Three space-separated integers, respectively: N, M, and X 
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

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

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

求目标点到图中的其他点来回的最小值。

Dijkstra直接求来回的距离，然后比较求出最小值。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;const int MAX = 100005;
int edge[1005][1005];
int vist[1005],vist2[1005],minidis1[1005][1005],minidis2[1005][1005];
int N,M,X;void init()
{int i,j;for(i=1;i<=N;i++){for(j=1;j<=N;j++){if(j==i){edge[i][j]=0;minidis1[i][j]=0;minidis2[i][j]=0;}else{edge[i][j]=-1;minidis1[i][j]=MAX;minidis2[i][j]=MAX;}}}for(i=1;i<=N;i++){vist[i]=0;vist2[i]=0;}
}void dijkstra(int i)
{int j,k;int position=i;int position2=i;vist[position]=1;vist2[position]=1;minidis1[i][position]=0;minidis2[position][i]=0;for(j=1;j<=N-1;j++)//一共要添加进num-1个点{for(k=1;k<=N;k++){if(vist[k]==0 && edge[position][k]!=-1 && minidis1[i][position]+edge[position][k] < minidis1[i][k])//新填入的点更新minidis{minidis1[i][k]=minidis1[i][position]+edge[position][k];}if(vist2[k]==0 && edge[k][position2]!=-1 && minidis2[position2][i]+edge[k][position2] < minidis2[k][i])//新填入的点更新minidis{minidis2[k][i]=minidis2[position2][i]+edge[k][position2];}}int min_value=MAX,min_pos=0;int min_value2=MAX,min_pos2=0;for(k=1;k<=N;k++){if(vist[k]==0 && minidis1[i][k]<min_value)//比较出最小的那一个作为新添入的店{min_value = minidis1[i][k];min_pos = k;}if(vist2[k]==0 && minidis2[k][i]<min_value2)//比较出最小的那一个作为新添入的店{min_value2 = minidis2[k][i];min_pos2 = k;}}vist[min_pos]=1;position=min_pos;vist2[min_pos2]=1;position2=min_pos2;}}int main()
{int i;cin>>N>>M>>X;init();int temp1,temp2,temp3;for(i=1;i<=M;i++){cin>>temp1>>temp2>>temp3;edge[temp1][temp2]=temp3;}memset(vist,0,sizeof(vist));memset(vist2,0,sizeof(vist2));dijkstra(X);int ans=-1;for(i=1;i<=N;i++){if(i==X)continue;ans=max(ans,minidis1[X][i]+minidis2[i][X]);}cout<<ans<<endl;return 0;
}

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