[C/C++] 1046 Shortest Distance (20 分)

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 D₁D₂⋯ 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

---

#include

int main()
{
	int n,m;
	scanf("%d",&n);
	int dst[n+1];
	int sum = 0;
	for(int i=1; i<=n; i++){
		scanf("%d",&dst[i]); 
		sum += dst[i];
	} 
	scanf("%d",&m);
	int start,end,short_path;
	for(int i=0; i<m; i++){
		scanf("%d%d",&start, &end);
		if( start>end ){
			int t = start;
			start = end;
			end = t;
		}
		int one = 0,another = 0;
		for(int j=start;j<end;j++){
			one += dst[j];
		}
		another = sum - one;
		printf("%d\n",one<another?one:another);
	}
		
	return 0;
}

^ 测试点2运行超时 (极端情况，10⁵ * 10⁴ = 10⁹),不经过预处理很容易超时
2.

#include

int main()
{
	int n,m;
	scanf("%d",&n);
	int dst[n] = {0};//dst[i]表示1到i+1的路长 
	int sum = 0,d;
	for(int i=1; i<=n; i++){
		scanf("%d",&d); 
		dst[i] = dst[i-1] + d;
		sum += d; 
	} 
	scanf("%d",&m);
	int start,end;
	for(int i=0; i<m; i++){
		scanf("%d%d",&start, &end);
		if( start>end ){
			int t = start;
			start = end;
			end = t;
		}
		int one;
		one = dst[end-1] - dst[start-1];
		printf("%d\n",one<another?one:sum - one);
	}
		
	return 0;
}