【luogu2850】[USACO06DEC]虫洞Wormholes

【luogu2850】[USACO06DEC]虫洞Wormholes

(​​ 题目描述

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’s 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.

John在他的农场中闲逛时发现了许多虫洞。虫洞可以看作一条十分奇特的有向边，并可以使你返回到过去的一个时刻（相对你进入虫洞之前）。John的每个农场有M条小路（无向边)连接着N （从1..N标号）块地，并有W个虫洞。其中1<=N<=500,1<=M<=2500,1<=W<=200。 现在John想借助这些虫洞来回到过去（出发时刻之前），请你告诉他能办到吗。 John将向你提供F(1<=F<=5)个农场的地图。没有小路会耗费你超过10000秒的时间，当然也没有虫洞回帮你回到超过10000秒以前。

输入输出格式

输入格式：

Line 1: A single integer, F. F farm descriptions follow.

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.

输出格式：

Lines 1..F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).

输入输出样例

输入样例#1：

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 8 输出样例#1：

NO YES 说明

For farm 1, FJ cannot travel back in time.

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.

题解：dfs-spfa判断负环，当然bfs也可以，因为数据量小

今天再次将变量名写错。我。浪费很多时间

#include #include #define N 550#define inf 0x7fffffffinline int read(){ int x=0;char ch=getchar(); while (ch<'0'||ch>'9') ch=getchar(); while (ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();} return x;}int num,h[N],n,m,w; struct node{ int x,y,z,next;}data[6000];inline void insert1(int x,int y,int z){ data[++num].x=x;data[num].y=y;data[num].z=z;data[num].next=h[x];h[x]=num; //data[++num].x=y;data[num].y=x;data[num].z=z;data[num].next=h[y];h[y]=num;}bool flag,visit[N];int f[N];void spfa(int x){ if (visit[x]==true){ visit[x]=false;flag=true;return; } visit[x]=true; for (int i=h[x];i;i=data[i].next){ int y=data[i].y,z=data[i].z; if (f[x]+z

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