Layout
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 9098   Accepted: 4347

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.

Source

[Submit]   [Go Back]   [Status]  
[Discuss]

Home Page   Go
Back
  To top

#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
#define MAXN 1010
#define MAXM 1000000+10
#define INF 10000000+10
int head[MAXN],dist[MAXN],used[MAXN],vis[MAXN];
int n,x,y,cnt;
struct node
{
int u,v,val;
int next;
}edge[MAXM];
void init()
{
memset(head,-1,sizeof(head));
cnt=0;
}
void add(int u,int v,int val)
{
node E={u,v,val,head[u]};
edge[cnt]=E;
head[u]=cnt++;
}
void getmap()
{
for(int i=1;i<n;i++)
add(i+1,i,0);
int a,b,c;
while(x--)
{
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);
}
while(y--)
{
scanf("%d%d%d",&a,&b,&c);
add(b,a,-c);
}
}
void SPFA()
{
queue<int> Q;
for(int i = 1; i <= n; i++)
{
dist[i] = i==1 ? 0 : INF;
vis[i] = false;
used[i] = 0;
}
// memset(vis,0,sizeof(vis));
// memset(dist,INF,sizeof(dist));
// memset(used,0,sizeof(used));
// dist[1]=0;
used[1] = 1;
vis[1] = 1;
Q.push(1);
while(!Q.empty())
{
int u = Q.front();
Q.pop();
vis[u] = 0;
for(int i = head[u]; i != -1; i = edge[i].next)
{
node E = edge[i];
if(dist[E.v] > dist[u] + E.val)
{
dist[E.v] = dist[u] + E.val;
if(!vis[E.v])
{
vis[E.v] = 1;
used[E.v]++;
if(used[E.v] > n)
{
printf("-1\n");
return ;
}
Q.push(E.v);
}
}
}
}
if(dist[n] == INF)
printf("-2\n");
else
printf("%d\n", dist[n]);
}
int main()
{
while(scanf("%d%d%d",&n,&x,&y)!=EOF)
{
init();
getmap();
SPFA();
}
return 0;
}

poj--3169--Layout(简单差分约束)的更多相关文章

  1. (简单) POJ 3169 Layout,差分约束+SPFA。

    Description Like everyone else, cows like to stand close to their friends when queuing for feed. FJ ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. POJ 3169 Layout 【差分约束】+【spfa】

    <题目链接> 题目大意: 一些母牛按序号排成一条直线.有两种要求,A和B距离不得超过X,还有一种是C和D距离不得少于Y,问可能的最大距离.如果没有最大距离输出-1,如果1.n之间距离任意就 ...

  6. 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] - ...

  7. poj 3169 Layout(差分约束)

    Layout Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6549   Accepted: 3168 Descriptio ...

  8. POJ 3167 Layout(差分约束)

    题面 Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 ...

  9. POJ 3169 Layout (差分约束系统)

    Layout 题目链接: Rhttp://acm.hust.edu.cn/vjudge/contest/122685#problem/S Description Like everyone else, ...

  10. O - Layout(差分约束 + spfa)

    O - Layout(差分约束 + spfa) Like everyone else, cows like to stand close to their friends when queuing f ...

随机推荐

  1. SVO在ROS下的配置与运行

    最近在做实验的时候,需要配置SVO,下面讲讲其中的过程以及遇到的问题: 首先说明配置环境:Ubuntu 14.04 + ROS indigo,ROS的安装我参考了ROS的官网上给出的教程:http:/ ...

  2. [原创]C++带空格字符串的输入问题

    字符串一直是一个重点加难点,很多笔试面试都会涉及,带空格的字符串更是十分常见,现在对字符串的输入问题进行一下总结. C++用cin输入的时候会忽略空格以后的字符,比如 char a[100]; cin ...

  3. DataTable的Select()方法

    DataRow[] partno = dtPack.Select("PK_SOHEAD = " + pk_sohead + " AND PART_NO = '" ...

  4. 使用postman模拟登录请求

    Cookie 可以写在Headers里面

  5. 『MicroPython』Hello uPy

    官网买了几乎全套.一路曲折:7月10号下单,13号发货,14号法兰克福过关,23号到北京,25号到上海,27号到沪C:沪C邮局投3次未果,中彩票一样终于打通了投递部电话才在次日28号“妥投”:又因出差 ...

  6. window窗口操作

    打开新窗口 window.open([url],[窗口名称],[参数字符串]) window.open("http://baidu.com","_balnk", ...

  7. 优动漫PAINT新建文件

    在优动漫PAINT软件中展开任何一项操作之前,都需要新建或打开图形文件.新建文件之后,用户可根据自己的需求进行相应的设置,这样将大大节省后期制作的时间! 在优动漫PAINT中新建图形文件的方法: 方法 ...

  8. CorelDRAW最高立返500元!还剩30个名额!速抢!

    由于上月CDR X7返利活动收获众多好评 本月官方继续将活动进行到底! 而此次活动不但有上月意犹未尽的CDR X7版,更增加了CDR X6.CDR 2017以及可望不可即的CDR 2018版,可谓是优 ...

  9. 配置H3C交换机ftp服务

    配置H3C交换机ftp服务,用于与交换机进行文件上传.下载,常用于更新程序上传及配置备份文件下载. 准备工作:三层设备(路由器.三层交换机等)至少一个接口配置IP,二层交换机需配置一个处于UP状态的v ...

  10. spring boot (一)

    spring boot 启动注解  @SpringBootApplication @Target(ElementType.TYPE) @Retention(RetentionPolicy.RUNTIM ...