poj3259 Wormholes【最短路-bellman-负环】
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.
Input
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 backT seconds.
Output
Sample Input
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
Sample Output
NO
YES
Hint
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.
题意:走一条路会花费时间 走虫洞会时间倒流 问能不能走回起点的时候时间倒流
思路:其实就是判断有没有环 有环的话
看题的时候看了半天不知道起点到底是哪一个点
感觉现在bellman写的还挺顺手的了
一个加边的addedge函数 一个判断松弛的relax函数
然后外面一个for循环遍历n-1次 里面的for循环松弛每一条边
再对每一条边判断能不能松弛 能就说明有环
有环其实就说明这个路径上有负的
不然干吗要不停的重复走???
只有可以不断减小才会形成环
WA了一发是因为加边的时候的cnt没有初始化
代码:
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<cstring>
#include<queue>
#include<stack>
#define inf 0x3f3f3f3f
using namespace std;
int f, n, m, w, cnt;
struct edge{
int s, e, t;
}path[5205];
long long d[505];
void addedge(int s, int e, int t)
{
path[cnt].s = s;
path[cnt].e = e;
path[cnt].t = t;
cnt++;
}
bool relax(int j)
{
if(d[path[j].e] > d[path[j].s] + path[j].t){
d[path[j].e] = d[path[j].s] + path[j].t;
return true;
}
return false;
}
bool bellman(int sec)
{
memset(d, inf, sizeof(d));
d[sec] = 0;
for(int i = 0; i < n - 1; i++){
bool flag = false;
for(int j = 0; j < cnt; j++){
if(relax(j)) flag = true;
}
if(!flag) return false;
}
for(int i = 0; i < cnt; i++){
if(relax(i)) return true;
}
return false;
}
int main()
{
cin>>f;
while(f--){
cin>>n>>m>>w;
cnt = 0;
for(int i = 0; i < m; i++){
int a, b, c;
cin>>a>>b>>c;
addedge(a, b, c);
addedge(b, a, c);
}
for(int i = 0; i < w; i++){
int a, b, c;
cin>>a>>b>>c;
addedge(a, b, -c);
}
if(bellman(1))
cout<<"YES\n";
else
cout<<"NO\n";
}
return 0;
}
poj3259 Wormholes【最短路-bellman-负环】的更多相关文章
- POJ3259 Wormholes 【spfa判负环】
题目链接:http://poj.org/problem?id=3259 Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submis ...
- POJ-3259 Wormholes (ballman_ford 判负环)
ballman_ford 是对单源点到任意点最短路的处理方法(可以含负权边). 对所有边进行n-1次循环,(n为点得个数),如果此时源点到这条边终点的距离 大于 源点到这条边起点的距离加上路得权值就进 ...
- POJ--3259 Wormholes (SPFA判负环)
题目电波 3259 Wormholes #include<iostream> #include<cstring> #include<algorithm> #in ...
- poj-3259 Wormholes(无向、负权、最短路之负环判断)
http://poj.org/problem?id=3259 Description While exploring his many farms, Farmer John has discovere ...
- POJ 3259 Wormholes 虫洞(负权最短路,负环)
题意: 给一个混合图,求判断是否有负环的存在,若有,输出YES,否则NO.有重边. 思路: 这是spfa的功能范围.一个点入队列超过n次就是有负环了.因为是混合图,所以当你跑一次spfa时发现没有负环 ...
- POJ3259:Wormholes(spfa判负环)
Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 68097 Accepted: 25374 题目链接: ...
- POJ:3259-Wormholes(最短路判断负环)
Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 58153 Accepted: 21747 Descripti ...
- Poj 3259 Wormholes(spfa判负环)
Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 42366 Accepted: 15560 传送门 Descr ...
- poj 3259 Wormholes【spfa判断负环】
Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 36729 Accepted: 13444 Descr ...
- (简单) POJ 3259 Wormholes,SPFA判断负环。
Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes ...
随机推荐
- iOS js与objective-c的交互(转)
在写 JavaScript 的时候,可以使用一个叫做 window 的对象,像是我们想要从现在的网页跳到另外一个网页的时候,就会去修改 window.location.href 的位置:在我们的 Ob ...
- PHP 免费获取手机号码归属地
一.淘宝网API API地址: http://tcc.taobao.com/cc/json/mobile_tel_segment.htm?tel=15850781443 参数: tel:手机号码 返回 ...
- SpringMVC由浅入深day01_8springmvc和mybatis整合
8 springmvc和mybatis整合 为了更好的学习 springmvc和mybatis整合开发的方法,需要将springmvc和mybatis进行整合. 整合目标:控制层采用springmvc ...
- web.xml 整合 SpringMVC
啦啦啦 <context-param> <param-name>defaultHtmlEscape</param-name> <param-value> ...
- Java NIO原理 图文分析及代码实现
Java NIO原理图文分析及代码实现 前言: 最近在分析hadoop的RPC(Remote Procedure Call Protocol ,远程过程调用协议,它是一种通过网络从远程计算机程序上请 ...
- HDOJ 4276 The Ghost Blows Light(树形DP)
Problem Description My name is Hu Bayi, robing an ancient tomb in Tibet. The tomb consists of N room ...
- nginx: [error] open() "/run/nginx.pid" failed (2: No such file or directory)
当你执行sudo nginx -s reload时出现nginx: [error] open() "/run/nginx.pid" failed (2: No such file ...
- cocos2d-x 3.0 在C++中调用lua函数(2)
个人觉得3.0里面, 在C++下面调用lua函数很不方便, 所以就扩展了一个类, 继承自LuaStack, 代码和使用方式如下: #ifndef __CC_LUA_STACKEX_H_ #define ...
- Selenium 前进和后退
我们平常使用浏览器时都有前进和后退功能, Selenium 也可以完成这个操作,它使用 back() 方法后退,使用 forward() 方法前进 from selenium import webdr ...
- windows之IP地址(一)
服务器: 顾名思义,是服务我们上网的机器.结构组成上和普通的PC结构相似,装了服务器版的系统后都可以作为服务器使用,但前者可靠性高.服务器是网络这个网上的中间节点,供每台终端进行访问.数据信息保存,有 ...