1046. Shortest Distance (20)

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1.建立一个数组dp[i]，记录0到i之间的距离和

2.定义sum，记录全程的路程和

3.求i，j的最短距离时，利用ans=dp[j]-[i]+num[i]-num[j]，ans2=sum-ans,min(ans,ans2)即为答案

AC代码：

#include <iostream>#include <stdio.h>#include <vector>#include <stack>#include <algorithm>#include <memory.h>#include <map>#include <set>#include "limits.h"using namespace std;int main(void){        int n;    scanf("%d",&n);    int *num=new int[n];    int *dp=new int[n];    memset(dp, 0, sizeof(dp));    int sum=0;    for(int i=0;i<n;i++)    {        scanf("%d",&num[i]);        if(i==0) dp[i]=num[i];        else dp[i]=dp[i-1]+num[i];        sum+=num[i];    }    int m;    scanf("%d",&m);    for(int i=0;i<m;i++)    {        int a,b;        scanf("%d %d",&a,&b);        if(a>b) swap(a,b);        int tmp=dp[b-1]-dp[a-1]+num[a-1]-num[b-1];        int tmp2=sum-tmp;        printf("%d\n",min(tmp2,tmp));    }    return 0;}

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