[kuangbin甯︿綘椋瀅涓撻鍥?鏈€鐭矾缁冧範 F - Wormholes 锛堝垽鏂礋鐜級
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鏍囩锛?a href='http://www.mamicode.com/so/1/wormholes' title='wormholes'>wormholes wing possible starting move can 绌烘牸 single sel
F - Wormholes
棰樼洰閾炬帴锛?a href="https://vjudge.net/contest/66569#problem/F" target="_blank">https://vjudge.net/contest/66569#problem/F
棰樼洰锛?/p>
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ鈥榮 farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.
2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8Sample Output
NO YESHint
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.
Input
绗?琛岋細涓€涓暣鏁癋琛ㄧず鎺ヤ笅鏉ヤ細鏈塅涓啘鍦鸿鏄庛€?姣忎釜鍐滃満绗竴琛岋細鍒嗗埆鏄笁涓┖鏍奸殧寮€鐨勬暣鏁帮細N锛孧鍜學 绗?琛屽埌M+1琛岋細涓変釜绌烘牸鍒嗗紑鐨勬暟瀛楋紙S锛孍锛孴锛夋弿杩帮紝鍒嗗埆涓猴細闇€瑕乀绉掕蛋杩嘢鍜孍涔嬮棿鐨勫弻鍚戣矾寰勩€備袱涓尯鍩熷彲鑳界敱涓€涓互涓婄殑璺緞鏉ヨ繛鎺ャ€?绗琈 +2鍒癕+ W+1琛岋細涓変釜绌烘牸鍒嗗紑鐨勬暟瀛楋紙S锛孍锛孴锛夋弿杩拌櫕娲烇紝鎻忚堪鍗曞悜璺緞锛孲鍒癊涓斿洖婧疶绉掋€?br />
Output
F琛岋紝姣忚浠ｈ〃涓€涓啘鍦?姣忎釜鍐滃満鍗曠嫭鐨勪竴琛岋紝” YES”琛ㄧず鑳芥弧瓒宠姹傦紝”NO”琛ㄧず涓嶈兘婊¤冻瑕佹眰銆?/div>
// // Created by hanyu on 2019/7/19. // #include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; const int maxn=2e5+7; #define MAX 0x3f3f3f3f int book[maxn],cnt[maxn],d[maxn],head[maxn]; int n,m,w; int pos; struct Node int s; int e; int t; node[maxn]; void add(int s,int e,int t) node[pos].s=e; node[pos].e=t; node[pos].t=head[s]; head[s]=pos++; bool spfa(int start) queue<int>qu; qu.push(start); book[start]=1; d[start]=0; while(!qu.empty()) int now=qu.front(); qu.pop(); book[now]=0; for(int i=head[now];i!=-1;i=node[i].t) int ss=node[i].s; int ee=node[i].e; if(d[ss]>d[now]+ee) d[ss]=d[now]+ee; if(!book[ss]) qu.push(ss); book[ss]=1; cnt[ss]++; if(cnt[ss]>=n) return true;//鍒ゆ柇璐熺幆 return false; int main() int T; scanf("%d",&T); while(T--) memset(book,0,sizeof(book)); memset(cnt,0,sizeof(cnt)); memset(d,MAX,sizeof(d)); memset(head,-1,sizeof(head)); memset(node,0,sizeof(node)); scanf("%d%d%d",&n,&m,&w); int s,e,t; for(int i=1;i<=m;i++) scanf("%d%d%d",&s,&e,&t); add(s,e,t); add(e,s,t); for(int i=1;i<=w;i++) scanf("%d%d%d",&s,&e,&t); add(s,e,-t); cnt[1]=1; pos=0; if(spfa(1)) printf("YES\n"); else printf("NO\n"); return 0;
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