洛谷 P1564 膜拜

题目传送门

解题思路:

把甲的崇拜者设为-1,乙的崇拜者设为1,求前缀和,f[i]表示到第i个人所需的最少机房数.

AC代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

int n,m,a[2501],j,y,f[2501],sum[2501];

int main() {
    memset(f,0x3f3f3f,sizeof(f));
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n; i++) {
        scanf("%d",&a[i]);
        if(a[i] == 2) sum[i] = sum[i-1] - 1;
        else sum[i] = sum[i-1] + 1; 
    }
    f[0] = 0;
    for(int i = 1;i <= n; i++)
        for(int j = 1;j <= i; j++)
            if(abs(sum[i] - sum[j-1]) == i - j + 1 || abs(sum[i] - sum[j-1]) <= m)
            //这一段全相同或两种人数相差不超过m 
                f[i] = min(f[i],f[j-1] + 1);
    printf("%d",f[n]);
    return 0;
}