C/C++[Shortest Distance]

题目描述
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.
一个包含N个节点的环，求任意两个节点之间的最短距离．
输入
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 … DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN 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.
每个测试用例：
- 第一行包含一个整数N,同行空格间隔，是N个整数，代表环上N个相邻节点之间的距离．
- 第二行也是一个整数M，表示下面要输入几个节点对
- M个节点对．
输出
For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.
两节点之间的最短距离（顺时针，逆时针）
样例输入
5 1 2 4 14 9
3
1 3
2 5
4 1
样例输出
3
10
7

a[N+1],存储相邻节点之间的距离，从a[1]开始，a[1]:节点１到节点２的距离．a[i]:distance(i, i+1)
这个简单图是一个环，所以任意两节点之间之存在两条路径，顺时针路径和逆时针路径，
１．计算出环总长度sum;
２．计算任意节点之间的顺时针距离dist_1;
３．逆时针距离　＝　sum - dist_1 = dist_2;
４．比较dist_1和dist_2的大小;

#include <stdio.h>
int main() {int N;while (scanf("%d",&N) != EOF){int a[N+1];int sum = 0;for (int i = 1; i < N+1; i++) {scanf("%d",&a[i]);sum += a[i];}int m;scanf("%d",&m);while(m--) {int c, d,dist_1=0, dist_2;scanf("%d %d",&c, &d);int temp;if (c > d) {temp = d;d = c;c = temp;}for(int i = c; i < d; i++)dist_1 += a[i];dist_2 = sum - dist_1;if(dist_1 < dist_2) printf("%d\n",dist_1);else printf("%d\n",dist_2);}}return 0;
}

解法二:上个解法时间复杂度较高．
方法二:不同地方是用dis存储从起始点到第i+１点的距离
顺时针距离　＝　dis[r-1] - dis[l-1]
逆时针距离　＝　sum - 顺时针距离

#include <iostream>
#include <vector>
using namespace std;
int main() {int n;cin >> n;vector<int> dis(n+1);int sum = 0, l, r, m;for (int i = 1; i <= n; i++) {int temp;cin >>temp;sum += temp;dis[i] = sum;}cin >> m;while(m--) {scanf("%d %d",&l, &r);if (l > r) swap(l, r);int dist = dis[r - 1] - dis[l -1];cout <<min(dist, sum - dist) <<endl;}return 0;
}

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