poj 3268 Silver Cow Party

Silver Cow Party

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 12776 Accepted: 5705

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

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.

Source

USACO 2007 February Silver

读完题目首先想到的是floyd算法，尝试着写了一遍，果断超时，然后就改成了Dijkstra算法，注意从X到其他点的的时候用正向图Dijkstra一遍，然后用反向图在Dijkstra一遍，遍历找出最大值即可。果断贴上代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 1005#define MAXN 1010#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int mp1[MAXN][MAXN],mp2[MAXN][MAXN];int dist1[MAXN],dist2[MAXN];int visit[MAXN];int N,M,X;void Dijkstra(int mp[][MAXN],int dist[]){    int i,j,mi,now;    memset(dist,INF,sizeof(dist));    memset(visit,0,sizeof(visit));    for (i=1;i<=N;i++)        dist[i]=mp[X][i];    visit[X]=1;    dist[X]=0;    for (i=1;i<=N;i++)    {        mi=INF;        now=-1;        for (j=1;j<=N;j++)        {            if (!visit[j]&&mi>dist[j])            {                now=j;                mi=dist[j];            }        }        visit[now]=1;        for (j=1;j<=N;j++)            if (!visit[j]&&mp[now][j]<INF&&dist[j]>dist[now]+mp[now][j])                dist[j]=dist[now]+mp[now][j];    }}int main(){    int i,j;    while (~scanf("%d%d%d",&N,&M,&X))    {        memset(mp1,INF,sizeof(mp1));        memset(mp2,INF,sizeof(mp2));        int u,v,w;        for (i=0;i<M;i++)        {            scanf("%d%d%d",&u,&v,&w);            mp1[u][v]=w;            mp2[v][u]=w;        }        memset(dist1,INF,sizeof(dist1));        memset(dist2,INF,sizeof(dist2));        Dijkstra(mp1,dist1);        Dijkstra(mp2,dist2);        int ans=-1;        for (i=1;i<=N;i++)            if (i!=X)        {            if (dist1[i]+dist2[i]>ans)                ans=dist1[i]+dist2[i];        }        printf("%d\n",ans);    }    return 0;}/*4 8 21 2 41 3 21 4 72 1 12 3 53 1 23 4 44 2 3*/