2 条题解

  • 0
    @ 2025-10-8 17:00:45

    by hansang:

    #include <bits/stdc++.h>
    using namespace std;
    const int N = 2e3 + 10;
    typedef long long LL;
    char s[N]; LL f[N][N];
    struct node{LL x, y, mn;} a[N];
    int main(){
        int n, m; scanf("%d%d", &m, &n);
        scanf("%s", s + 1);
        for(int i = 1; i <= m; i++){
            char ss[5]; scanf("%s", ss);
            int t = ss[0] - 'a';
            scanf("%lld%lld", &a[t].x, &a[t].y);
            a[t].mn = min(a[t].x, a[t].y);
        }
        memset(f, 0x3f, sizeof(f));
        for(int i = n; i >= 1; i--){
            f[i][i] = 0;
            for(int j = i + 1; j <= n; j++){
                int t1 = s[i] - 'a', t2 = s[j] - 'a';
                f[i][j] = min(f[i+1][j] + a[t1].mn, f[i][j-1] + a[t2].mn);
                if(s[i] == s[j]){
                    if(i + 1 == j) f[i][j] = 0;
                    else f[i][j] = min(f[i][j], f[i+1][j-1]);
                }
            }
        }
        printf("%lld\n", f[1][n]);
        return 0;
    }
    
    • 0
      @ 2025-10-8 17:00:35

      by hansang:

      #include<bits/stdc++.h>
      using namespace std;
      const int N=2e3+10;
      typedef long long LL;
      char s[N]; LL f[N][N];
      struct node{LL x, y, mn;} a[N];
      int main(){
          int n, m; scanf("%d%d", &m, &n);
          scanf("%s", s+1);
          for(int i=1; i<=m; i++){
              char ss[5]; scanf("%s", ss);
              int t=ss[0]-'a';
              scanf("%lld%lld", &a[t].x, &a[t].y);
              a[t].mn=min(a[t].x, a[t].y);
          }
          memset(f, 0x3f, sizeof(f));
          for(int i=n; i>=1; i--){
              f[i][i]=0;
              for(int j=i+1; j<=n; j++){
                  int t1=s[i]-'a', t2=s[j]-'a';
                  f[i][j]=min(f[i+1][j]+a[t1].mn, f[i][j-1]+a[t2].mn);
                  if(s[i]==s[j]){
                      if(i+1==j) f[i][j]=0;
                      else f[i][j]=min(f[i][j], f[i+1][j-1]);
                  }
              }
          }
          printf("%lld\n", f[1][n]);
          return 0;
      }
      • 1

      USACO(110)动态规划(区间型)2:修改回文P2890 [USACO07OPEN] Cheapest Palindrome

      信息

      ID
      2301
      时间
      1000ms
      内存
      128MiB
      难度
      7
      标签
      递交数
      15
      已通过
      9
      上传者