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
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.
思路：思路很好想，就是正反建圖，分別從x點(diǎn)跑最短路就可以，最后加起來求最大值。樣例很坑，有重邊。poj上，g++交的話，必須得用堆優(yōu)化才可以過，否則超時(shí)（也有可能是我代碼的問題，總之注意一下吧）。
代碼如下：

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue> #define ll long long #define inf 0x3f3f3f3f using namespace std;const int maxx=1e3+100; int mp1[maxx][maxx],mp2[maxx][maxx]; struct node{int pos;int v;node(){}node(int a,int b){pos=a,v=b;}bool operator <(const node &a)const{return v>a.v;} }; int dis1[maxx],dis2[maxx]; int vis[maxx]; int n,m,x,tot1,tot2;inline void init() {memset(dis1,inf,sizeof(dis1));memset(dis2,inf,sizeof(dis2));memset(mp1,inf,sizeof(mp1));memset(mp2,inf,sizeof(mp2));tot1=tot2=0; }inline void Dijkstra1(int u) {memset(vis,0,sizeof(vis));priority_queue<node> p;p.push(node(u,0));dis1[u]=0;while(p.size()){node r=p.top();p.pop();int v=r.pos;if(vis[v]) continue;vis[v]=1;for(int i=1;i<=n;i++){if(mp1[v][i]!=inf){int to=i;int w=mp1[v][i];if(dis1[to]>dis1[v]+w) {dis1[to]=dis1[v]+w;p.push(node(to,dis1[to]));}}}} } inline void Dijkstra2(int u) {memset(vis,0,sizeof(vis));priority_queue<node> p;p.push(node(u,0));dis2[u]=0;while(p.size()){node r=p.top();p.pop();int v=r.pos;if(vis[v]) continue;vis[v]=1;for(int i=1;i<=n;i++){if(mp2[v][i]!=inf){int to=i;int w=mp2[v][i];if(dis2[to]>dis2[v]+w) {dis2[to]=dis2[v]+w;p.push(node(to,dis2[to]));}}}} } int main() {while(scanf("%d%d%d",&n,&m,&x)!=EOF){init();int a,b,c;for(int i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);mp1[a][b]=mp2[b][a]=min(mp1[a][b],c);}for(int i=1;i<=n;i++) mp1[i][i]=mp2[i][i]=0;Dijkstra1(x);Dijkstra2(x);int _max=0;for(int i=1;i<=n;i++) _max=max(_max,dis1[i]+dis2[i]);printf("%d\n",_max);}return 0; }

努力加油a啊，(^o)/~

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