原题传送门
这个是双关键字的单调队列
记 H H H为1, G G G为-1,统计前缀和
暴力 d p i = m i n ( d p j + ( s u m i > = s u m j ) ) ( i − j < = m ) dp_i=min(dp_j+(sum_i>=sum_j))(i-j<=m) dpi=min(dpj+(sumi>=sumj))(i−j<=m)
用单调队列维护,如何判断两个数在队列中的先后?
对于两个数 p , q p,q p,q
- d p p ! = d p q dp_p!=dp_q dpp!=dpq, d p dp dp值小的那个优先
- d p p = d p q dp_p=dp_q dpp=dpq, s u m sum sum值小的那个优先,因为更可能达到+0的情况
Code:
#include <bits/stdc++.h>
#define maxn 300010
using namespace std;
int sum[maxn], dp[maxn], n, m, q[maxn];
int get(){
char c = getchar();
for (; c != 'H' && c != 'G'; c = getchar());
return c == 'H' ? 1 : -1;
}
int main(){
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + get(), dp[i] = 1e9;
int h = 1, t = 1;
for (int i = 1; i <= n; ++i){
while (h <= t && i - q[h] > m) ++h;
if (h <= t) dp[i] = dp[q[h]] + (sum[i] <= sum[q[h]]);
while (h <= t && (dp[i] < dp[q[t]] || dp[i] == dp[q[t]] && sum[i] <= sum[q[t]])) --t;
q[++t] = i;
}
printf("%d\n", dp[n]);
return 0;
}