UVa1619 Feel Good (标记,枚举)

Posted 2020-08-09

链接：http://vjudge.net/problem/UVA-1619

 

分析：首先计算前缀和，后面会用到。设每个i都属于一个满足下述条件的最大区间（因为题意说是正整数序列，区间当然越大越好），a[i]为区间中的最小值，然后预处理出以i为轴向左和向右延伸的最远位置L[i]和R[i]来标记这个区间，最后就是O（n）的枚举。

 

 1 #include <cstdio>
 2 
 3 const int maxn = 1e5+5;
 4 
 5 int n, a[maxn], L[maxn], R[maxn];
 6 long long prefix_sum[maxn];
 7 
 8 int main() {
 9     int T = 0;
10     while (scanf("%d", &n) == 1) {
11         if (T++) printf("\n");
12         a[0] = -1; a[n+1] = -1; prefix_sum[0] = 0;
13         for (int i = 1; i <= n; i++) {
14             scanf("%d", &a[i]);
15             prefix_sum[i] = prefix_sum[i-1] + a[i];
16             L[i] = i; R[i] = i;
17         }
18         for (int i = 1; i <= n; i++)
19             while (a[i] <= a[L[i]-1]) L[i] = L[L[i]-1];
20         for (int i = n; i >= 1; i--)
21             while (a[i] <= a[R[i]+1]) R[i] = R[R[i]+1];
22         long long ans = a[1] * a[1];
23         int l = 1, r = 1;
24         for (int i = 1; i <= n; i++) {
25             long long v = (prefix_sum[R[i]] - prefix_sum[L[i]-1]) * a[i];
26             if (v > ans || (v == ans && r-l > R[i]-L[i])) {
27                 ans = v;
28                 l = L[i], r = R[i];
29             }
30         }
31         printf("%lld\n", ans);
32         printf("%d %d\n", l, r);
33     }
34     return 0;
35 }