1046 Shortest Distance (20point(s)) Easy only once *数组预处理问题

Posted 2021-03-16 songlinxuan

基本思想：

对于这个环形的正权值队列来说，完全可以从第一个节点计数，用1~i和1~j节点的距离来计算i~j节点的距离；

 

注意点：

1.对于高次个数，遍历不靠谱，找机会打表和优化结构；

2.对于正负权值要注意；

 

超时代码：

 1 #include<iostream>
 2 #include<stdlib.h>
 3 #include<stdio.h>
 4 #include<vector> 
 5 #include<string>
 6 #include<math.h>
 7 #include<algorithm>
 8 using namespace std;
 9 using std::vector; 
10 const int maxn = 100009;
11 int dis[maxn];
12 int main() {
13     int n;
14     scanf("%d", &n);
15     for (int i = 0; i < n; i++)
16         scanf("%d", &dis[i]);
17     int m;
18     scanf("%d", &m);
19     for (int i = 0; i < m; i++) {
20         int a, b;
21         scanf("%d %d", &a, &b);
22         a--;
23         b--;
24         int d1=0,d2=0;
25         int in = a;
26         while (in != b) {
27             d1 += dis[in];
28             in = (in + 1) % n;
29         }
30         in = a;
31         while (in != b&&d2<d1) {
32             in = ((in - 1) + n) % n;
33             d2 += dis[in];
34         }
35         printf("%d
", min(d1, d2));
36     }
37     system("pause");
38     return 0;
39 }

 

预处理代码：

 1 #include<iostream>
 2 #include<stdlib.h>
 3 #include<stdio.h>
 4 #include<vector> 
 5 #include<string>
 6 #include<math.h>
 7 #include<algorithm>
 8 using namespace std;
 9 using std::vector; 
10 const int maxn = 100009;
11 int dis[maxn] = {0};
12 int dis_pre[maxn];
13 int main() {
14     int n;
15     int sum = 0;
16     scanf("%d", &n);
17     for (int i = 1; i <= n; i++) {
18         scanf("%d", &dis[i]);
19         dis_pre[i+1] = dis[i] + dis_pre[i];
20         sum += dis[i];
21     }
22     int m;
23     scanf("%d", &m);
24     for (int i = 0; i < m; i++) {
25         int a, b;
26         scanf("%d %d", &a, &b);
27         int d = abs(dis_pre[b] - dis_pre[a]);
28         printf("%d
", min(d,sum-d));
29     }
30     system("pause");
31     return 0;
32 }