UVa1437 String painter

emmm...好难的一道dp

其实本质是两次dp.第一次求出都不相等时候的最优解, 第二次在求出两串有相同时的更优的解.

https://blog.csdn.net/hndu__lz/article/details/52167981

这位讲的不错, 可以去看看.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 1e2 + 20;

char s[MAXN], t[MAXN];
int f[MAXN][MAXN], g[MAXN];
int N;

int main()
{
    while(scanf("%s
%s", s, t) == 2)
    {
        N = strlen(s);
        for(int i = 0; i < N; i++) f[i][i] = 1;
            
        for(int len = 1; len < N; len++)
            for(int l = 0; l + len < N; l++){
                int r = l + len;
                f[l][r] = f[l + 1][r] + 1;
                
                for(int k = l + 1; k <= r; k++)
                    if(t[l] == t[k]) f[l][r] = min(f[l][r], f[l + 1][k] + f[k + 1][r]);
            }
            
        for(int r = 0; r < N; r++)
        {
            g[r] = f[0][r];
            if(s[r] == t[r]) g[r] = (r == 0) ? 0 : g[r - 1];
            else 
                for(int l = 0; l < r; l++) g[r] = min(g[r], g[l] + f[l + 1][r]);
        }
        cout<<g[N - 1]<<endl;
    }
    return 0;
}