标签：line 题目 limit describe back silver art bsp hint
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 12494 | Accepted: 5568 |
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; roadi requires Ti (1 ≤ Ti ≤ 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
Output
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
Source
所以我们要枚举全部的可能性。找出当中的最大值。
巧妙地运用dijkstra算法，双向求出两次X->m的最短路径长然后相加即得到了m->X->m的最短路径。代码例如以下：#include<queue>
#include<vector>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct Edge
{
int to;
int dis;
Edge(int to, int dis){
this -> to = to;
this -> dis = dis;
}
};
typedef pair<int,int>P;
int a,b,c;
int N,M,X;
int d1[1005],d2[1005];
vector<Edge> G1[1005];
vector<Edge> G2[1005];
void dijkstra(int s,int d[],vector<Edge> G[])
{
priority_queue<P,vector<P>,greater<P> >q;
d[s]=0;
q.push(P(0,s));
while(q.size())
{
P p=q.top();
q.pop();
int v=p.second;
for(int i=0;i<G[v].size();i++)
{
Edge& e=G[v][i];
if(d[e.to]>d[v]+e.dis)
{
d[e.to]=d[v]+e.dis;
q.push(P(d[e.to],e.to));
}
}
}
}
int main()
{
memset(d1,0x5f,sizeof(d1));
memset(d2,0x5f,sizeof(d2));
scanf("%d%d%d",&N,&M,&X);
for(int i=1;i<=M;i++)
{
scanf("%d%d%d",&a,&b,&c);
G1[a].push_back(Edge(b,c));
G2[b].push_back(Edge(a,c));
}
dijkstra(X,d1,G1);
dijkstra(X,d2,G2);
int small_max=-1;
for(int i=1;i<=N;i++)
{
if(i==X) continue;
small_max=max(small_max,d1[i]+d2[i]);
}
cout<<small_max<<endl;
}
标签：line 题目 limit describe back silver art bsp hint
原文地址：http://www.cnblogs.com/yutingliuyl/p/7137910.html
