HDU5163Taking Bus

Taking Bus

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1209    Accepted Submission(s): 383

Problem Description

Bestland has a very long road and there are n bus station along the road, which are numbered 1 to n from left to right. There are m persons wanting to take the bus to some other station. You task is to find the time needed for each person. Note: All the other information you need is below. Please read the statment carefully.

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤60), indicating the number of test cases. For each test case: The first line contains two integers n and m (2≤n,m≤105), indicating the number of bus stations and number of people. In the next line, there are n−1 integers, d1,d2,…,dn−1 (1≤di≤109). The i-th integer means the distance between bus station i and i+1 is di (1≤i<n). In the next m lines, each contains two integers xi and yi (1≤xi,yi≤n,xi≠yi), which means i-th person is in bus station xi and wants goto bus station yi. (1≤i≤m)

What else you should know is that for the i-th person, the bus starts at bus station ((i−1) mod n)+1 and drives to right. When the bus arrives at station n, it will turn around and drive from right to left. Similarly, When the bus arrives at station 1, it will turn around and drive from left to right. You can assume that the bus drives one meter per second. And you should only consider the time that the bus drives and ignore the others.

 

Output

For each person, you should output one integer which is the minimum time needed before arriving bus station yi.

 

Sample Input

1
7 3
2 3 4 3 4 5
1 7
4 5
5 4

 

BEST CODE 里面的题目，看不懂题意的可以去看中文版的：http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=563&pid=1002

不是很难，也是水题；理解题意就可以知道，只要分三种情况，就可以做了；

给出代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int sta[100005];//存取每个公交站点之间的距离
long long suml[100005];//计算每个公交站点到第一个公交站点的距离
long long sum[100005];//存取每一组数据所花费的最少时间；
int main()
{
	int T,n,m,start,end;
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d%d",&n,&m);
		long long num=0;
		for (int i = 2; i <= n; i++)
		{
			scanf("%lld", &sta[i]);
			num += sta[i];
			suml[i] = num;             //处理每一个公交站点到第一个公交站点的距离；
		}
		for (int count = 1; count <= m; count++)  //计数，
		{
			scanf("%d%d",&start,&end);
			int Staion = (count - 1) % n + 1;//第count个人要坐车时，公交车的起始站点位置；
//最好画图模拟下，这是人的终点站点标号>起始站点标号，并且公交车的起始站点标号<=人的起始站点标号的情况；
			if (end>start&&Staion<=start)   
			{
				
				sum[count] = suml[end] - suml[Staion];
			}
//人的终点站点标号>起始站点标号，并且公交车的起始站点标号 > 人的起始站点标号的情况；
			else if (end>start&&Staion > start)
			{
				sum[count] = 2 * suml[n] - suml[Staion] + suml[end];
			}
//人的终点站点标号<起始站点标号，并且公交车的起始站点标号 > 人的起始站点标号或者公交车的起始站点标号<=人的起始站点标号的情况；；
			else if (start > end)
			{
				sum[count] = 2 * suml[n] - suml[Staion] - suml[end];
			}
		}
		for (int i = 1; i <= m; i++)
			printf("%lld\n", sum[i]);
	}
	return 0;
}