Educational Codeforces Round 57 (Rated for Div. 2) D dp

https://codeforces.com/contest/1096/problem/D

题意

给一个串s,删掉一个字符的代价为a[i],问使得s的子串不含"hard"的最小代价

题解

• 定义\(dp[i][j]\)为到第i位下一个将要匹配j的最小代价
  • \(若s[i]==t[j]\)
    • 删掉:\(min(dp[i+1][j],dp[i][j]+a[i])\)
    • 不删,若j<3:\(min(dp[i+1][j+1],dp[i][j])\)
  • \(若s[i]!=t[j]\)
    • 不用删:\(min(dp[i+1][j],dp[i][j])\)
• \(ans=min(dp[n][i]),0 \leq i \leq 3\)

代码

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
#define MAXN 100005
using namespace std;
ll dp[MAXN][4],a[MAXN],ans;
int n;string s;
int main(){
    cin>>n>>s;
    memset(dp,inf,sizeof(dp));
    for(int i=0;i<n;i++)scanf("%lld",&a[i]);
    string t="hard";
    dp[0][0]=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<4;j++){
           if(dp[i][j]==inf)continue;
           if(s[i]==t[j]){
              dp[i+1][j]=min(dp[i+1][j],dp[i][j]+a[i]);
              if(j<3)dp[i+1][j+1]=min(dp[i+1][j+1],dp[i][j]);
           }else{
               dp[i+1][j]=min(dp[i+1][j],dp[i][j]);
           }
        }
    }
    ans=1e18;
    for(int i=0;i<4;i++)ans=min(ans,dp[n][i]);
    cout<<ans;
}