1046 Shortest Distance-PAT甲级

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 [3,10​5​​]), followed by N integer distances D​1​​ D​2​​ ⋯ 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 (≤10​4​​), 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 10​7​​.

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

解题思路：dst[i]数组表示结点i+1到结点1的距离，dst[n]表示整个环的距离，求任意两点i与j的距离可以表示成

res=min(abs(dst[j-1]-dst[i-1]),abs(dst[n]-abs(dst[j-1]-dst[i-1)) 的形式

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=100005;
int dst[N];
int main(){
	memset(dst,0,sizeof(dst));
	int n,m;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&dst[i]);
		dst[i]+=dst[i-1];
	}
	scanf("%d",&m);
	int u,v,res;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&u,&v);
		res=min(abs(dst[v-1]-dst[u-1]),abs(dst[n]-abs(dst[v-1]-dst[u-1])));
		printf("%d\n",res);
	}
	return 0;
}

标签：Distance,PAT,1046,exits,distance,int,dst,between,each
来源： https://blog.csdn.net/amf12345/article/details/100171865