Shortest Path

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1494    Accepted Submission(s): 476

Problem Description
There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1 to n and there is an edge with unit length between i and i+1 (1≤i<n). To make the graph more interesting, someone adds three more edges to the graph. The length of each new edge is 1.

You are given the graph and several queries about the shortest path between some pairs of vertices.

 
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains two integer n and m (1≤n,m≤105) -- the number of vertices and the number of queries. The next line contains 6 integers a1,b1,a2,b2,a3,b3 (1≤a1,a2,a3,b1,b2,b3≤n), separated by a space, denoting the new added three edges are (a1,b1), (a2,b2), (a3,b3).

In the next m lines, each contains two integers si and ti (1≤si,ti≤n), denoting a query.

The sum of values of m in all test cases doesn't exceed 106.

 
Output
For each test cases, output an integer S=(∑i=1mi⋅zi) mod (109+7), where zi is the answer for i-th query.
 
Sample Input
1
10 2
2 4 5 7 8 10
1 5
3 1
 
Sample Output
7
 
Source
 
Recommend
wange2014   |   We have carefully selected several similar problems for you:  5659 5658 5657 5655 5654 
 最近在学习最短路问题,floyd写法。
void floyd(){
    for(int k = 1; k<=6; k++)
        for(int i = 1; i<=6; i++)
            for(int j = 1; j<=6; j++)
                d[i][j] = min(d[i][j],d[i][k]+d[k][j]);
}
 #include <iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn = ;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+;
int d[][];
int a[];
void floyd(){
for(int k = ; k<=; k++)
for(int i = ; i<=; i++)
for(int j = ; j<=; j++)
d[i][j] = min(d[i][j],d[i][k]+d[k][j]);
}
void solve(){
int t,n,m;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
for(int i = ; i<=; i++) scanf("%d",&a[i]);
for(int i = ; i<=; i++){
for(int j = ; j<=; j++){
d[i][j] = abs(a[i] - a[j]);
}
}
for(int i = ; i<=; i+=) d[i][i+] = d[i+][i] = ;
floyd();
long long sum = ; for(int k = ; k<=m; k++){
int x,y;
scanf("%d%d",&x,&y);
int ans = abs(x-y);
for(int i = ; i<=; i++)
for(int j = ; j<=; j++)
ans = min(ans,(abs(x-a[i])+d[i][j]+abs(y-a[j])));
sum = (sum + (long long)ans*k%mod)%mod;
}
printf("%I64d\n",sum);
}
}
int main()
{
solve();
return ;
}

卷珠帘

Shortest Path的更多相关文章

  1. hdu-----(2807)The Shortest Path(矩阵+Floyd)

    The Shortest Path Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others ...

  2. zoj 2760 How Many Shortest Path 最大流

    题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1760 Given a weighted directed graph ...

  3. The Shortest Path in Nya Graph

    Problem Description This is a very easy problem, your task is just calculate el camino mas corto en ...

  4. hdu 3631 Shortest Path(Floyd)

    题目链接:pid=3631" style="font-size:18px">http://acm.hdu.edu.cn/showproblem.php?pid=36 ...

  5. Shortest Path(思维,dfs)

    Shortest Path  Accepts: 40  Submissions: 610  Time Limit: 4000/2000 MS (Java/Others)  Memory Limit: ...

  6. (中等) HDU 4725 The Shortest Path in Nya Graph,Dijkstra+加点。

    Description This is a very easy problem, your task is just calculate el camino mas corto en un grafi ...

  7. 【ZOJ2760】How Many Shortest Path

    How Many Shortest Path 标签: 网络流 描述 Given a weighted directed graph, we define the shortest path as th ...

  8. [Swift]LeetCode847. 访问所有节点的最短路径 | Shortest Path Visiting All Nodes

    An undirected, connected graph of N nodes (labeled 0, 1, 2, ..., N-1) is given as graph. graph.lengt ...

  9. OSPF(Open Shortest Path First)

    1.概述 路由协议OSPF全称为Open Shortest Path First,也就开放的最短路径优先协议,因为OSPF是由IETF开发的,所以所有厂商都可以用. OSPF的流量使用IP协议号. O ...

随机推荐

  1. LeetCode OJ 96. Unique Binary Search Trees

    Given n, how many structurally unique BST's (binary search trees) that store values 1...n? For examp ...

  2. Python之路:爬虫之urllib库的基本使用和高级使用

    关于爬虫自己一直在看,所以时间太慢,这才第二更,有等不及的小伙伴可以慢慢的品尝了,在看下面的之前,建议先把上一章看一下.以下是关于python的Urllib的基础和高级用法. 1.如何扒下一个网站,用 ...

  3. android动态添加TextView或者ImageView

    动态添加 text1=new TextView(this); text1.setText("动态添加"); ((LinearLayout) this.findViewById(R. ...

  4. SpringSecurity自定义过滤器

    applicationContext-security.xml: <beans:beans xmlns="http://www.springframework.org/schema/s ...

  5. 关于video.js

    网址:http://www.cnblogs.com/webenh/p/5815741.html

  6. asp.net html table to DataTable

    添加引用 http://htmlagilitypack.codeplex.com/downloads/get/437941 protected void Export(string content,s ...

  7. Inno Setup入门(十六)——Inno Setup类参考(2)

    分类: Install Setup 2013-02-02 11:28 815人阅读 评论(0) 收藏 举报 这里将接着在前面的基础上介绍如何在自定义页面上添加按钮.按钮属于Tbutton类,该类继承自 ...

  8. mongodb状态

    基本信息 spock:PRIMARY>db.serverStatus() { "host" :"h6.corp.yongche.org", //主机名 & ...

  9. zookeeper集群实例

    zookeeper是什么 Zookeeper,一种分布式应用的协作服务,是Google的Chubby一个开源的实现,是Hadoop的分布式协调服务,它包含一个简单的原语集,应用于分布式应用的协作服务, ...

  10. RS232 DB9 公头 母头 串口引脚定义

    转自:http://hi.baidu.com/zhy_myspace/item/ad1ab0f112a4da1ad7ff8cf9 DB9 公头 母头 串口引脚定义 1.RS-232端(DB9母头/孔型 ...