甲级PAT 1046 Shortest Distance （20 分）（动态规划）

1046 Shortest Distance （20 分）

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

题目要求: 

有N个出站口，他们之间是呈环形连通。给出每两个出站口之间的距离，求两个出站口之间的最小距离。

第一行分别是，N，D1,D2……Di，表示第i个出站口到第i+1个出站口的距离.

解题思路：

由于这里是一个环形，因此从一个出站口到另一个出站口无非两条路径，一个是顺时针，一个是逆时针。两个方向的距离之和为整个环的长度。

用一个数组dis[i]来存储第i+1个出口到第一个出口的距离，这样任意两个出口i,j之间的距离可用dis[j - 1] - dis[i - 1] （j > i）来表示。此时另一条路径的距离用dis[N] - (dis[j - 1] - dis[i - 1] )。最后输出两条路径距离小的那个就可以了。

对于j < i的情况可以首先交换两个的位置用上述公式，因为都是要求两条路径，先求哪个都一样。

随手画了一个比较丑的图。。凑合理解。

完整代码：

#include<iostream>
#include<cstring>
using namespace std;
int main(){
	int i,N,M,d,a,b,r1,r2;
	int dis[100010];
	cin >> N;
	memset(dis,0,sizeof(dis));
	for(i = 1; i <= N; i++){
		cin >> d;
		dis[i] = dis[i-1] + d;
	}
	cin >> M;
	for(i = 1; i <= M; i++){
		cin >> a >> b;
		if(a > b) swap(a,b);
		r1 = dis[b-1] - dis[a-1];
		r2 = dis[N] - r1;
		cout<<min(r1,r2)<<endl;
	}
	return 0;
}