Silver Cow Party

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 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 farmA_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 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

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

題目的意思是：從出發(fā)點(diǎn)到達(dá)目的地X。再從X返回到出發(fā)點(diǎn)的最短路徑中的最大值（由于出發(fā)點(diǎn)沒有固定，也就是能夠：1->X->1, 2->X->2等）。

所以我們要枚舉全部的可能性。找出當(dāng)中的最大值。

巧妙地運(yùn)用dijkstra算法，雙向求出兩次X->m的最短路徑長然后相加即得到了m->X->m的最短路徑。代碼例如以下：
#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; }

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