Codeforces Gym - 101617A Ducks in a Row [DP]
2017-12-06 17:10
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题意:给一个只包含‘D'和’G‘的字符串,每次操作可以把一个区间内的’D‘变’G‘,’G‘变’D‘,问最少需要几次,能出现至少k个长度大于等于n的连续的‘D’。
题解:可以定义dp[i][j][2]表示i位置为’D‘或’G‘往前有有j个长度大于等于n的连续的'D'的最小步数。
首先当前位置可以可以从上一个位置的情况转移过来,然后假设要在当前位置向前形成n个’D‘,那么需要的步数就是这一段中0的个数,需要注意的是,有步数减少的情况:
AC代码:
题解:可以定义dp[i][j][2]表示i位置为’D‘或’G‘往前有有j个长度大于等于n的连续的'D'的最小步数。
首先当前位置可以可以从上一个位置的情况转移过来,然后假设要在当前位置向前形成n个’D‘,那么需要的步数就是这一段中0的个数,需要注意的是,有步数减少的情况:
dp[i][j][1]=min(dp[i][j][1],min(dp[i-1][j][0]+1-(s[i-1]=='D'?1:0),dp[i-1][j][1]+1-(s[i-1]=='G'?1:0))); if(i-len<0)continue; if(j)dp[i][j][1]=min(dp[i][j][1],dp[i-len][j-1][0]+need[i-len+1][i]-(s[i-len]=='D'&&s[i-len+1]=='G'?1:0));
AC代码:
#include<stdio.h>
#include<string.h>
#include<iostream>
#define inf 10000000
#define N 2005
using namespace std;
int dp
[2],need
;
char s
;
int main()
{
int len,cot;
scanf("%d%d",&len,&cot);
scanf("%s",s+1);
int l=strlen(s+1);
for(int i=1;i<=l;i++)
{
if(s[i]=='G')need[i][i]=1;
for(int j=i+1;j<=l;j++)
{
need[i][j]=need[i][j-1];
if(s[j-1]!=s[j]&&s[j]=='G')need[i][j]++;
}
}
s[0]='G';
for(int i=0;i<N;i++)
for(int j=0;j<N;j++)
dp[i][j][0]=dp[i][j][1]=inf;
dp[0][0][0]=0;
dp[0][0][1]=1;
int ans=inf;
for(int i=1;i<=l;i++)
for(int j=0;j<=cot;j++)
{
if(s[i]=='G')
{
dp[i][j][0]=min(dp[i][j][0],min(dp[i-1][j][0],dp[i-1][j][1]));
dp[i][j][1]=min(dp[i][j][1],min(dp[i-1][j][0]+1-(s[i-1]=='D'?1:0),dp[i-1][j][1]+1-(s[i-1]=='G'?1:0))); if(i-len<0)continue; if(j)dp[i][j][1]=min(dp[i][j][1],dp[i-len][j-1][0]+need[i-len+1][i]-(s[i-len]=='D'&&s[i-len+1]=='G'?1:0));
}
else
{
dp[i][j][0]=min(dp[i][j][0],min(dp[i-1][j][0]+1-(s[i-1]=='D'?1:0),dp[i-1][j][1]+1-(s[i-1]=='G'?1:0)));
dp[i][j][1]=min(dp[i][j][1],min(dp[i-1][j][0],dp[i-1][j][1]));
if(i-len<0)continue;
if(j)dp[i][j][1]=min(dp[i][j][1],dp[i-len][j-1][0]+need[i-len+1][i]-(s[i-len]=='D'&&s[i-len+1]=='G'?1:0));
}
if(j==cot)ans=min(ans,min(dp[i][j][0],dp[i][j][1]));
}
if(ans==inf)printf("-1\n");
else printf("%d\n",ans);
}
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