poj 3268 Silver Cow Party(单源最短路径Dijkstra·最小环)

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题目：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=11757

Silver Cow Party

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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 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

Sample Output

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.

终点给出，是一个定点。所有的牛犊都会到达该点并且由通过不同的道路回到原点。由此想到单源最短路径，把终点看作源点S，但是怎么回去呢？看了一个大神的博客，矩阵转置（第一次自己转置都写错了||-_-)。效果：(1,2) (2,7) (7,4) --> (2,1) (7,2) (4,7) means (4,7) (7,2) (2,1) 起点和终点的地位互换！这样再一次用Dijkstra或者spfa等算法求出回家的最短路径就构成了环路。
关于矩阵转置举个例子：
for(int i=0;i<4;i++){
for(int j=0;j<=i;j++){
swap(a.m[i][j],a.m[j][i]);
}
}
不是：
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
swap(a.m[i][j],a.m[j][i]);
}
}
来看看效果：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
struct matrie{int m[4][4];
};
matrie a={
1,2,3,4,
7,3,5,5,
6,6,6,6,
0,0,0,0
};
void show(){cout<<"show: \n";for(int i=0;i<4;i++){for(int j=0;j<4;j++){cout<<a.m[i][j]<<" ";}cout<<endl;}
}
void trans(){for(int i=0;i<4;i++){for(int j=0;j<=i;j++){swap(a.m[i][j],a.m[j][i]);cout<<i<<" "<<j<<": "<<endl;show();}}
}int main()
{trans();return 0;
}

0 0:
show:
1 2 3 4
7 3 5 5
6 6 6 6
0 0 0 0
1 0:
show:
1 7 3 4
2 3 5 5
6 6 6 6
0 0 0 0
1 1:
show:
1 7 3 4
2 3 5 5
6 6 6 6
0 0 0 0
2 0:
show:
1 7 6 4
2 3 5 5
3 6 6 6
0 0 0 0
2 1:
show:
1 7 6 4
2 3 6 5
3 5 6 6
0 0 0 0
2 2:
show:
1 7 6 4
2 3 6 5
3 5 6 6
0 0 0 0
3 0:
show:
1 7 6 0
2 3 6 5
3 5 6 6
4 0 0 0
3 1:
show:
1 7 6 0
2 3 6 0
3 5 6 6
4 5 0 0
3 2:
show:
1 7 6 0
2 3 6 0
3 5 6 0
4 5 6 0
3 3:
show:
1 7 6 0
2 3 6 0
3 5 6 0
4 5 6 0
是的，就是这样。
Dijkstra：基于贪心的思想。不足之处在于不能有负边，用于求解单源最短路径问题。这里没有负边问题，所以我就用它吧。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1005,INF=0x3f3f3f3f;
int dis1[maxn],dis2[maxn],map[maxn][maxn],pre[maxn];
int n,m,s;
bool p[maxn];
void Dijkstra(int dis[]){memset(p,0,sizeof(p));for(int i=1;i<=n;i++){if(i!=s){ dis[i]=map[s][i]; pre[i]=s; }else pre[i]=0;}dis[s]=0;p[s]=1;for(int i=1;i<n;i++){  //循环n-1次，求s到其他n-1个点的最短距离int minm=INF,k=0;for(int j=1;j<=n;j++){ //寻找最近的点if(p[j]==0&&dis[j]<minm){minm=dis[j];k=j;}}if(k==0) return;  //没有可以拓展的点p[k]=1;for(int j=1;j<=n;j++){  if(p[j]==0&&map[k][j]!=INF&&dis[j]>dis[k]+map[k][j]){ //利用最近的点更新其他点的距离dis[j]=dis[k]+map[k][j];pre[j]=k;}}}
}
void trans(){for(int i=1;i<=n;i++){for(int j=1;j<=i;j++){swap(map[i][j],map[j][i]);}}
}
int main()
{//freopen("cin.txt","r",stdin);while(cin>>n>>m>>s){int a,b,c;memset(map,0x3f,sizeof(map));for(int i=1;i<=n;i++) map[i][i]=0;while(m--){scanf("%d%d%d",&a,&b,&c);if(a!=b&&map[a][b]>c)map[a][b]=c;}Dijkstra(dis1);trans();Dijkstra(dis2);int ans=0;for(int i=1;i<=n;i++){if(i!=s)ans=max(ans,dis1[i]+dis2[i]);}printf("%d\n",ans);}return 0;
}

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