PAT 1046. Shortest Distance (20)【卡的是超时】

最新推荐文章于 2021-09-12 23:15:53 发布

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1046. Shortest Distance (20)

时间限制

100 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (in [3, 10⁵]), followed by N integer distances D₁ D₂ ... D_N, where D_i is the distance between the i-th and the (i+1)-st exits, and D_N is between the N-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integer M (<=10⁴), with M lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 10⁷.

Output Specification:

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

Sample Input:

5 1 2 4 14 9
3
1 3
2 5
4 1

Sample Output:

3
10
7

提交代码

/*
	此题卡的是超时。。必须在输入时就预处理好 1 到各点的距离。。另外，初始化dist[i]为全0，默认是负无穷大。。每次用变量之前要初始化。。。。。。
*/
#include<stdio.h>
#include<string.h>
const int MAX=100000;
int min(int a,int b)
{
	return a<b?a:b;
}
int max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int i,n,m,b,c,tmp1,tmp2;
	//freopen("G:\\in.txt","r",stdin);
	//freopen("G:\\our.txt","w",stdout);
	while(scanf("%d",&n)!=EOF){
		int ans,a[MAX],dist[MAX],sum=0;
		for(i=0;i<n;i++)
			dist[i]=0;
		for(i=1;i<=n;i++){
			scanf("%d",&a[i]);   //scanf()里的变量别忘了&。
			sum+=a[i];
			dist[i+1]=sum; //dist[i]保存1到i的距离。dist[n]保存一整圈的距离。
		}
		scanf("%d",&m);
		while(m--){
			ans=0;
			scanf("%d%d",&b,&c);
			tmp1=min(b,c);tmp2=max(b,c); //注意啦：给出的两个数不知道前面的大还是后面的大。
			ans=dist[tmp2]-dist[tmp1];
			if(sum-ans<ans)
				ans=sum-ans;
			printf("%d\n",ans);
		}
	}
	return 0;
}