1046. Shortest Distance (20)

时间限制

100 ms

内存限制

65536 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 14 931 32 54 1

Sample Output:

3107

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;const int maxn=100001;int ans[maxn],dis[maxn];#if 0int main(){int n;memset(ans,0,sizeof(ans));int sumnum=0;scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&ans[i]);sumnum+=ans[i];}int M;scanf("%d",&M);int a,b,minnum,sum,x,y;while(M--){minnum=0;sum=0;scanf("%d %d",&a,&b);if(a>b){x=a;y=b;}if(a<b){x=b;y=a;}for(int i=y;i<x;i++){ //这里会导致第3个case超时sum+=ans[i];}minnum=sumnum-sum;if(sum<minnum)   minnum=sum;printf("%d\n",minnum);}//system("pause");return 0;}#endifint main(){int sum=0,n;scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d",&ans[i]);sum+=ans[i];dis[i]=sum;}int M,minnum;scanf("%d",&M);while(M--){int a,b,tmp;minnum=0;scanf("%d%d",&a,&b);if(a<b){tmp=a;a=b;b=tmp;}minnum=dis[a-1]-dis[b-1];minnum=min(minnum,sum-minnum);printf("%d\n",minnum);}//system("pause");return 0;}