POJ

题目链接：http://poj.org/problem?id=3268点击打开链接

Silver Cow Party

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 23722 Accepted: 10830

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.

对于这道题 要求来回 如果暴力肯定超时

可以反向思维 将其他点到达目标点看作是反向的目标点到达其他点

这样一来两遍dijkstra就行 一遍以正方向 一遍反方向

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>#include<queue>#include<map>#include<math.h>#include<limits.h>#include<vector>using namespace std;int n,m,x;struct xjy{    int num;    int dis;    bool operator < (const xjy &r) const    {        return dis>r.dis;    }};struct remember{    int mid1;    int mid2;    int mid3;};vector<remember> ss;xjy dis[1111];xjy mmap[1111][1111];int book[1111];int aans[1111];priority_queue<xjy>q;void dijkstra(int begin,int end){    xjy mid;    dis[begin].dis=0;    dis[begin].num=begin;    mid=dis[begin];    q.push(mid);    while(!q.empty())    {        mid=q.top();        q.pop();        for(int i=1;i<=n;i++)        {            if(!book[i]&&dis[i].dis>=mmap[mid.num][i].dis+dis[mid.num].dis&&mmap[mid.num][i].dis<INT_MAX)            {                if(dis[i].dis==mmap[mid.num][i].dis+dis[mid.num].dis)                    continue;                dis[i].dis=mmap[mid.num][i].dis+dis[mid.num].dis;                dis[i].num=i;                q.push(dis[i]);            }        }    }}int main(){        while(~scanf("%d%d%d",&n,&m,&x))    {        int ans=0;        for(int i=1;i<=1000;i++)            for(int j=1;j<=1000;j++)            {                mmap[i][j].dis=INT_MAX;            }        for(int i=1;i<=1000;i++)        {            dis[i].dis=INT_MAX;            book[i]=0;        }        for(int i=1;i<=m;i++)        {            remember mid;            scanf("%d%d%d",&mid.mid1,&mid.mid2,&mid.mid3);            if(mmap[mid.mid1][mid.mid2].dis>mid.mid3)            {                mmap[mid.mid1][mid.mid2].dis=mid.mid3;            }            ss.push_back(mid);        }        dijkstra(x,x);        for(int i=1;i<=n;i++)        {            aans[i]=dis[i].dis;        }                for(int i=1;i<=1000;i++)            for(int j=1;j<=1000;j++)            {                mmap[i][j].dis=INT_MAX;            }        for(int i=1;i<=1000;i++)        {            dis[i].dis=INT_MAX;            book[i]=0;        }        for(int i=0;i<m;i++)        {            remember mid=ss[i];            if(mmap[mid.mid2][mid.mid1].dis>mid.mid3)            {                mmap[mid.mid2][mid.mid1].dis=mid.mid3;            }        }        dijkstra(x,x);        ans=0;        for(int i=1;i<=n;i++)        {            aans[i]+=dis[i].dis;            ans=max(ans,aans[i]);        }        cout << ans << endl;    }}