Layout
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 11612   Accepted: 5550

Description

Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and since they can be rather pushy, it is possible that two or more cows can line up at exactly the same location (that is, if we think of each cow as being located at some coordinate on a number line, then it is possible for two or more cows to share the same coordinate).

Some cows like each other and want to be within a certain distance of each other in line. Some really dislike each other and want to be separated by at least a certain distance. A list of ML (1 <= ML <= 10,000) constraints describes which cows like each other and the maximum distance by which they may be separated; a subsequent list of MD constraints (1 <= MD <= 10,000) tells which cows dislike each other and the minimum distance by which they must be separated.

Your job is to compute, if possible, the maximum possible distance between cow 1 and cow N that satisfies the distance constraints.

Input

Line 1: Three space-separated integers: N, ML, and MD.

Lines 2..ML+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at most D (1 <= D <= 1,000,000) apart.

Lines ML+2..ML+MD+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at least D (1 <= D <= 1,000,000) apart.

Output

Line 1: A single integer. If no line-up is possible, output -1. If cows 1 and N can be arbitrarily far apart, output -2. Otherwise output the greatest possible distance between cows 1 and N.

Sample Input

4 2 1
1 3 10
2 4 20
2 3 3

Sample Output

27

Hint

Explanation of the sample:

There are 4 cows. Cows #1 and #3 must be no more than 10 units apart, cows #2 and #4 must be no more than 20 units apart, and cows #2 and #3 dislike each other and must be no fewer than 3 units apart.

The best layout, in terms of coordinates on a number line, is to put cow #1 at 0, cow #2 at 7, cow #3 at 10, and cow #4 at 27.

题意:有n头牛按编号顺序站一排,即每头牛都有一个一维坐标,可以相同。现在有一些牛之间有关系,关系好的a,b必须距离小于等于dl;关系不好的a,b必须距离大于等于dd。求牛1和牛n的最大距离。
思路:最短路问题:<u,v> d[u]+d>=d[v]。
n头牛按编号顺序站一排,则d[i+1]>=d[i],即编号大的牛的坐标大于等于编号小的牛。关系好的牛a,牛b,则d[a]+d>=d[b];关系不好的牛a,牛b,则d[a]+d<=d[b],即d[b]+(-d)>=d[a]。求约束下的最大距离。最短路也可以理解为约束下的最大解。
因为存在负权值,所以有可能存在负权值回路,所以dijkstra算法不能使用,直接使用ford算法。存在负权值回路输出-1,d[n]=inf输出-2,其他情况直接输出d[n]。
代码:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
const int maxn=1e5+,inf=0x3f3f3f3f,mod=1e9+;
const ll INF=1e13+;
struct edge
{
int from,to;
int cost;
};
int cou=;
edge es[maxn];
vector<edge>G[maxn];
int used[maxn];
priority_queue<P,vector<P>,greater<P> >que;
void addedge(int u,int v,int w)
{
cou++;
edge e;
e.from=u,e.to=v,e.cost=w;
es[cou].from=u,es[cou].to=v,es[cou].cost=w;
G[u].push_back(e);
}
int n,ml,md;
int al[maxn],bl[maxn],dl[maxn];
int ad[maxn],bd[maxn],dd[maxn];
int d[maxn];
void ford()
{
for(int i=; i<=n; i++) d[i]=inf;
d[]=;
for(int t=; t<n; t++)
{
for(int i=; i<n; i++)
if(d[i+]<inf) d[i]=min(d[i],d[i+]);
for(int i=; i<=ml; i++)
if(d[al[i]]<inf) d[bl[i]]=min(d[bl[i]],d[al[i]]+dl[i]);
for(int i=; i<=md; i++)
if(d[bd[i]]<inf) d[ad[i]]=min(d[ad[i]],d[bd[i]]-dd[i]);
}
if(d[]<) cout<<-<<endl;
else if(d[n]>=inf) cout<<-<<endl;
else cout<<d[n]<<endl;
}
int main()
{
int a,b,d;
scanf("%d%d%d",&n,&ml,&md);
for(int i=; i<n; i++) addedge(i+,i,);
for(int i=; i<=ml; i++)
scanf("%d%d%d",&al[i],&bl[i],&dl[i]);
for(int i=; i<=md; i++)
scanf("%d%d%d",&ad[i],&bd[i],&dd[i]);
ford();
return ;
}
/*
4 3 0
1 3 10
2 4 20
2 3 3
*/

最短路

POJ 3169.Layout 最短路的更多相关文章

  1. poj 3169 Layout (差分约束)

    3169 -- Layout 继续差分约束. 这题要判起点终点是否连通,并且要判负环,所以要用到spfa. 对于ML的边,要求两者之间距离要小于给定值,于是构建(a)->(b)=c的边.同理,对 ...

  2. poj 3169 Layout(线性差分约束,spfa:跑最短路+判断负环)

    Layout Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15349   Accepted: 7379 Descripti ...

  3. POJ 3169 Layout(差分约束+最短路)题解

    题意:有一串数字1~n,按顺序排序,给两种要求,一是给定u,v保证pos[v] - pos[u] <= w:二是给定u,v保证pos[v] - pos[u] >= w.求pos[n] - ...

  4. POJ 3169 Layout(差分约束啊)

    题目链接:http://poj.org/problem? id=3169 Description Like everyone else, cows like to stand close to the ...

  5. POJ 3169 Layout (spfa+差分约束)

    题目链接:http://poj.org/problem?id=3169 差分约束的解释:http://www.cnblogs.com/void/archive/2011/08/26/2153928.h ...

  6. POJ 3169 Layout (HDU 3592) 差分约束

    http://poj.org/problem?id=3169 http://acm.hdu.edu.cn/showproblem.php?pid=3592 题目大意: 一些母牛按序号排成一条直线.有两 ...

  7. poj 3169 Layout(差分约束+spfa)

    题目链接:http://poj.org/problem?id=3169 题意:n头牛编号为1到n,按照编号的顺序排成一列,每两头牛的之间的距离 >= 0.这些牛的距离存在着一些约束关系:1.有m ...

  8. poj 3169 Layout

    Layout Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8610   Accepted: 4147 Descriptio ...

  9. POJ 3169 Layout (spfa+差分约束)

    题目链接:http://poj.org/problem?id=3169 题目大意:n头牛,按编号1~n从左往右排列,可以多头牛站在同一个点,给出ml行条件,每行三个数a b c表示dis[b]-dis ...

随机推荐

  1. 微信小程序--分享报错(thirdScriptError Cannot read property 'from' of undefined;at pages/index/index page onShareAppMessage function TypeError: Cannot read property 'from' of undefined)

    分享功能: onShareAppMessage: function (res) { if (res.from === 'button') { // 来自页面内转发按钮 console.log(res. ...

  2. jenkins commande not found

    解决方法: http://www.geekcome.com/content-10-3868-1.html 1.控制台执行 echo $PATH 把输出的这句话复制 2.jenkins->系统管理 ...

  3. 2017面向对象程序设计(Java)第2周学习指导及要求(2017.8.28-2017.9.3)

    学习目标 继续适应老师教学方式的变化,能按照翻转课堂教学要求完成课前知识学习: 掌握Java Application程序结构: 掌握Java的数据类型与变量: 学会使用运算符构造各类表达式: 掌握输入 ...

  4. 设计table表格,用js设计偶数行和奇数行显示不同的颜色

    第一种:鼠标经过时table表格中的颜色根据奇偶行改变不同的颜色 <!DOCTYPE html> <html> <head> <meta charset=&q ...

  5. css中background-size的属性值

    length 设置背景图像的高度和宽度. 第一个值设置宽度,第二个值设置高度. 如果只设置一个值,则第二个值会被设置为 "auto". percentage 以父元素的百分比来设置 ...

  6. 第三章 列表(b)无序列表

  7. python os.path模块常用方法详解(转)

    转自:https://www.cnblogs.com/wuxie1989/p/5623435.html os.path模块主要用于文件的属性获取,在编程中经常用到,以下是该模块的几种常用方法.更多的方 ...

  8. Mac上反编译Android apk安装包

    什么是反编译 我们知道,Android的程序打包后会生成一个APK文件,这个文件可以直接安装到任何Android手机上,因此,反编译就是对这个APK进行反编译.Android的反编译分成两个部分: 一 ...

  9. ActiveMQ之整合spring

    ActiveMQ可以轻松的与Spring进行整合,Spring提供了一系列的接口类,非常的好用! 比如异步消息数据.异步发送邮件.异步消息查询等 <dependency> <grou ...

  10. PTA 7-9 旅游规划(SPFA)

    有了一张自驾旅游路线图,你会知道城市间的高速公路长度.以及该公路要收取的过路费.现在需要你写一个程序,帮助前来咨询的游客找一条出发地和目的地之间的最短路径.如果有若干条路径都是最短的,那么需要输出最便 ...