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
题目大意:给出一串数字(坐标从1开始), 这些数字围成一个圈,找出给出亮点的最小距离;
第一次试的时候,超时了,因为处理的不是很完善
解法:1.输入的时候,把所有的值加起来,计算的时候,只需要计算一个方向的距离就行, 另一个方向的距离用总的距离(即一个环的长度)来减就行;这种方法在最后一个测试点不能通过,因为每次计算都要重复的计算一些点的距离;
2.对一进行一些小的改进,把数组中的值,保存为距离第一个点的距离,这样在计算的时候,把两点离第一个点的距离想减,就得到两点的距离。 为了方便计算,添加一个下标为0的点。
描述点不是很清晰,还是看代码吧。
关键在于,在输入的时候,就对数据进行处理,减少循环次数
1 #include<iostream> 2 using namespace std; 3 int main(){ 4 int n, m, i, sum=0, *arr; 5 cin>>n; 6 arr = new int[n]; 7 for(i=0; i<n; i++){ 8 cin>>arr[i]; 9 sum += arr[i]; 10 } 11 cin>>m; 12 for(i=0; i<m; i++){ 13 int a, b, j, l=0; 14 cin>>a>>b; 15 if(a>b) swap(a, b); 16 for(j=a; j<b; j++) l += arr[j-1]; 17 l = l<(sum-l)?l:(sum-l); 18 cout<<l<<endl; 19 } 20 return 0; 21 }
修改之后点代码
1 #include<iostream> 2 using namespace std; 3 int main(){ 4 int n, m, i, sum=0, *arr; 5 cin>>n; 6 arr = new int[n+1]; 7 arr[0]=0; 8 for(i=1; i<=n; i++){ 9 cin>>arr[i]; 10 sum += arr[i]; 11 arr[i] += arr[i-1]; 12 } 13 cin>>m; 14 for(i=0; i<m; i++){ 15 int a, b, l=0; 16 cin>>a>>b; 17 if(a>b) swap(a, b); 18 l = arr[b-1]- arr[a-1]; 19 l = l<(sum-l)?l:(sum-l); 20 cout<<l<<endl; 21 } 22 return 0; 23 }