poj 3268 Silver Cow Party
来源:互联网 发布:淘宝 买药 编辑:程序博客网 时间:2024/04/30 12:50
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 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
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
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
这道题目其实就是正反向建两幅图,然后从源点x来两遍spfa,正向建图时,得到的是返回时的最短路,反向建图时得到的是去参加派对的最短路。
可惜这题一直WA啊,跪得要哭了,一直觉得代码没错啊。后来终于发现了,原来是队列开小了,因为我是自己手写队列,因此队头指针在不停地往后挪,因此尽管队列中最多顶点个数,但是其实下标会大于顶点总数的,这里要开大,如果不想用STL,又不确定开多大,那只能写循环队列了。
代码:
#include<cstdio>#include<iostream>#include<cstring>#define Maxv 1010#define Maxe 100010using namespace std;struct line{ int to,w; line(){} line(int t,int ww):to(t),w(ww){}}edge[Maxe],redge[Maxe];int head[Maxv],rhead[Maxv],nxt[Maxe],rnxt[Maxe];int q[Maxe],vis[Maxv],dist[Maxv],rdist[Maxv];const int inf=1<<30;void spfa(int u,int n,int *dist,int *head,int *nxt,line *edge){ for(int i=1;i<=n;i++) dist[i]=inf; dist[u]=0; memset(vis,0,sizeof vis); int s=0,e; q[e=0]=u; while(s<=e){ int v=q[s++]; vis[v]=0; for(int i=head[v];i!=-1;i=nxt[i]){ int r=edge[i].to,w=edge[i].w; if(dist[v]+w<dist[r]){ dist[r]=dist[v]+w; if(!vis[r]) q[++e]=r,vis[r]=1; } } }}int main(){ int n,m,x,fr,to,w; while(~scanf("%d%d%d",&n,&m,&x)){ memset(head,-1,sizeof head); memset(rhead,-1,sizeof rhead); for(int i=0;i<m;i++){ scanf("%d%d%d",&fr,&to,&w); edge[i]=line(to,w),nxt[i]=head[fr],head[fr]=i; redge[i]=line(fr,w),rnxt[i]=rhead[to],rhead[to]=i; } spfa(x,n,dist,head,nxt,edge); spfa(x,n,rdist,rhead,rnxt,redge); int ans=0; for(int i=1;i<=n;i++) ans=max(ans,dist[i]+rdist[i]); printf("%d\n",ans); }return 0;}
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