LG5202 「USACO2019JAN」Redistricting 动态规划+堆/单调队列优化

问题描述

LG5202

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题解

\[opt[i]=xx+(cnt[i]-cnt[yy]<=0) \]

发现\(cnt[i]-cnt[yy] <= 0\)只能有两种取值

于是直接堆优化即可

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\(\mathrm{Code}\)

#include<bits/stdc++.h>
using namespace std;

template <typename Tp>
void read(Tp &x){
	x=0;char ch=1;int fh;
	while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
	if(ch=='-'){
		fh=-1;ch=getchar();
	}
	else fh=1;
	while(ch>='0'&&ch<='9'){
		x=(x<<1)+(x<<3)+ch-'0';
		ch=getchar();
	}
	x*=fh;
}

const int maxn=300000+7;

int n,k;
int opt[maxn],cnt[maxn];
char s[maxn];

struct node{
	int val,pos;
	bool operator <(node a)const{
		return val==a.val?cnt[pos]>cnt[a.pos]:val>a.val;
	}
};

priority_queue<node>q;

int main(){
	read(n);read(k);cin>>(s+1);
	for(int i=1;i<=n;i++){
		if(s[i]=='H'){
			cnt[i]=cnt[i-1]+1;
		}
		else cnt[i]=cnt[i-1]-1;
	}
	q.push((node){0,0});
	for(int i=1;i<=n;i++){
		while(1){
			int x=q.top().pos;
			if(x>=i-k) break;
			q.pop();
		}
		int xx=q.top().val,yy=q.top().pos;
		opt[i]=xx+(cnt[i]-cnt[yy]<=0);
		q.push((node){opt[i],i});
	}
	printf("%d\n",opt[n]);
	return 0;
}