Silver Cow PartySilver Cow Party （最短路）

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概述

Silver Cow Party （最短路）

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 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

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: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

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

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.

解题报告

用Dijkstra走两次：以X为起点一个正向走一个逆向走

#include<stdio.h>
#include<queue>
#include<vector>
#define INF 0x3f3f3f
#define MAX_N 1002
using namespace std;
struct edge{int to,cost;};
typedef pair<int ,int> P;//f 距离 s 顶点
int V,E,X;
int d1[MAX_N],d2[MAX_N];
int map[MAX_N][MAX_N];
bool operator <(P x,P y){
return x.first>y.first;
}
void dijkstra(int s,vector<edge> *G,int *d){
priority_queue<P> que;
fill(d,d+V+1,INF);
d[s]=0;
que.push(P(0,s));
while(!que.empty()){
P p=que.top();que.pop();
int v=p.second;
if(d[v]<p.first) continue;
for(int i=0;i<G[v].size();i++){
edge e=G[v][i];
if(d[v]+e.cost<d[e.to]){
d[e.to]=d[v]+e.cost;
que.push(P(d[e.to],e.to));
}
}
}
}
int main()
{
while(~scanf("%d%d%d",&V,&E,&X)){
vector<edge> G1[V+1],G2[V+1];
int s;edge get;
while(E--){
scanf("%d%d%d",&s,&get.to,&get.cost);
G1[s].push_back(get);
int tmp=s;
s=get.to,get.to=tmp;
G2[s].push_back(get);
}
dijkstra(X,G1,d1);
dijkstra(X,G2,d2);
int max=0;
for(int i=1;i<=V;i++){
int tmp=d1[i]+d2[i];
if(max<tmp) max=tmp;
}
printf("%dn",max);
}
return 0;
}