CF2005C Lazy Narek

記錄dp的設計。一開始設計的是f[i][j]表示最後一個選i，匹配到j的最大值，然而這樣轉移是\(n^2\)的，題目要求\(n*m\). 設計成0，1揹包，考慮第i個選擇或者不選擇即可。

#include<bits/stdc++.h>
using namespace std;
const int N = 1e3 + 11;
int f[N][6];
int lef[N][6], val[N][6], to[N][6];
int len;
bool mark[N];
char s[N][N];
char S[] = {'n', 'a', 'r', 'e', 'k'};
void work(){
    int n, m;
    cin>>n>>m;
    for(int i = 1;i <= n; i++){
        scanf("%s", s[i]);
    }
    len = strlen(s[1]);
    for(int i = 0;i <= 4; i++){
        for(int j = 1;j <= n; j++){
            lef[j][i] = 0;
            int cnt = 0;
            int pos = i;
            for(int k = 0;k < len; k++){
                if(s[j][k] == S[pos]){
                    pos++;
                    if(pos == 5)cnt += 1, pos -= 5;
                }
                else {
                    bool flag = false;
                    for(int l = 0;l < 5; l++)if(s[j][k] == S[l])flag = true;
                    lef[j][i] += flag;
                }
            }
            to[j][i] = pos;
            val[j][i] = 5 * cnt;
        }
    }
    for(int i = 0;i <= n; i++){
        for(int j = 0;j < 5; j++)f[i][j] =  (i == 0 && j == 0) ? 0 : -1e8;
    }
    for(int i = 0;i < n; i++){
        for(int j = 0;j < 5; j++)f[i+1][j] = f[i][j];
        for(int j = 0;j < 5; j++){
            f[i+1][to[i+1][j]] = max(f[i+1][to[i+1][j]], f[i][j] + val[i+1][j] - lef[i+1][j]);
        }
    }
    int res = 0;
    for(int j = 0;j < 5; j++){
      res = max(res, f[n][j] - j);
    }
    cout<<res<<endl;
}
int main(){
    int T;
    cin>>T;
    while(T--)work();
    return 0;
}