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  • PTA(Advanced Level)1046.Shortest Distance

    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
    
    思路
    • 如果设起点、终点分别为x,y ,顺时针的距离就是dis(x,y),逆时针的距离就是dis(y,x),比较就好了。如果是对每一组测试点都手动模拟的话会TLE
    • 所以需要优化,这题的本质是区间和的比较,我们可以另开一个数组记录区间和,那么每次查询的时候只要直接相减就好了
    代码
    #include<bits/stdc++.h>
    using namespace std;
    int a[100010] = {0};
    int length[100010] = {0};
    int main()
    {
    	int n;
    	scanf("%d", &n);
    	int sum = 0;
    	for(int i=1;i<=n;i++)
    	{
    		scanf("%d", &a[i]);
    		sum += a[i];
    		length[i] = sum;
    	}
    	int m;
    	scanf("%d", &m);
    	int l, r;
    	int t; //暂时存储距离
    	while(m--)
    	{
    		scanf("%d %d", &l, &r);
    		if(l > r)	swap(l, r);
    		t = length[r-1] - length[l-1];
    		printf("%d
    ", min(t, sum - t));
    	}
    	return 0;
    }
    
    
    引用

    https://pintia.cn/problem-sets/994805342720868352/problems/994805435700199424

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  • 原文地址:https://www.cnblogs.com/MartinLwx/p/11650217.html
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