POJ3268Silver Cow Party（单源最短路径变形）

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题意：给n个顶点和m条边的有向图，输出每一个顶点到顶点X和顶点X到每一个顶点的最短路径之和最大的那个数。

思路：从顶点X到每一个顶点的最短路径不难想到直接套用Dijkstra算法即可得到，难的是每一个顶点到X的最短路，这里有一个巧妙的就是将原来的有向图各条边取反向，再在X做一次单源最短路，即为每个点到X点的最短路径，两者相加然后取最大即可

#include <cstdio>#include <queue>#include <cstring>#include <iostream>#include <cstdlib>#include <algorithm>#include <vector>#include <map>#include <string>#include <set>#include <ctime>#include <cmath>#include <cctype>using namespace std;#define maxn 1005#define LL long longconst int INF = 1<<29;int cas=1,T;int mapp[maxn][maxn];int n,m,x;int dijkstra(){   int mins,v;   int d[maxn];   int dd[maxn];   int vis[maxn];   for (int i = 1;i<=n;i++)   {   vis[i]=0;   d[i]=mapp[x][i];           //从x点出发   dd[i]=mapp[i][x];   }      for (int i = 1;i<=n;i++)   {       mins = INF;   for (int j = 1;j<=n;j++)     if (!vis[j] && d[j] < mins) { v=j; mins = d[j]; }   vis[v]=1;   for (int j = 1;j<=n;j++)   if (!vis[j] && d[j] > mapp[v][j]+d[v])   d[j]=mapp[v][j]+d[v];   }   memset(vis,0,sizeof(vis));   for (int i = 1;i<=n;i++)   {   mins = INF;   for (int j = 1;j<=n;j++)   if (!vis[j] && dd[j] < mins)   {   v=j;   mins = dd[j];   }   vis[v]=1;   for (int j = 1;j<=n;j++)   if (!vis[j] && dd[j] > mapp[j][v]+dd[v])   dd[j] = mapp[j][v]+dd[v];   }   int ans = -1;   for (int i = 1;i<=n;i++)       ans = max(ans,d[i]+dd[i]);   return ans;}int main(){scanf("%d%d%d",&n,&m,&x);for (int i = 1;i<=n;i++)for (int j = 1;j<=n;j++){if (i!=j)mapp[i][j]=INF;            elsemapp[i][j]=0;}for (int i = 1;i<=m;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);mapp[a][b]=c;}printf("%d\n",dijkstra());//freopen("in","r",stdin);//scanf("%d",&T);//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);return 0;}

题目

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

4 8 21 2 41 3 21 4 72 1 12 3 53 1 23 4 44 2 3

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.

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