luoguP2039 [AHOI2009]跳棋 巧妙的dp

(f[i])表示在第(i)个格子上弄一个棋子的最小代价,前后扫两遍dp后统计答案即可。

代码

#include<bits/stdc++.h>
using namespace std;
#define N 2007
#define ll long long
const ll inf=1e16;
ll f[N];
int tag[N];
int main()
{
	int n,i;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
		scanf("%d",&tag[i]);
	memset(f,0x3f,sizeof(f));
	for(i=2;i<=n;i++)
	{
		if(tag[i])f[i]=1;
		else f[i]=min(f[i],f[i-1]+f[i-2]);
	}
	for(i=n;i>=2;i--)
		f[i]=min(f[i],f[i+1]+f[i+2]);
	ll ans1=0,ans2=0;
	for(i=2;i<=n;i+=2)
	{
		if(f[i]<inf)ans2+=f[i];
		else ans1++;
	}
	printf("%lld
%lld
",ans1,ans2);
	return 0;
}

原文地址:https://www.cnblogs.com/lishuyu2003/p/11297125.html