HDU 2476 String painter（记忆化搜索， DP）

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const LL INF = 0xfffffff;
const LL maxn = 105;
int dp[maxn][maxn], ans[maxn];
char a[maxn], b[maxn];
int DFS(int L,int R)
{
    if(dp[L][R])
        return dp[L][R];
    if(L == R)
        return dp[L][R] = 1;
    if(L > R)
        return 0;
    dp[L][R] = DFS(L+1,R) + 1;
    for(int k=L+1; k<=R; k++)
    {
        if(b[L] == b[k])
            dp[L][R] = min(dp[L][R], DFS(L+1,k) + DFS(k+1,R) );
    }
    return dp[L][R];
}


int main()
{

    while(cin >> a >> b)
    {
        memset(dp, 0, sizeof(dp));
        int n = strlen(a);
        for(int i=0; i<n; i++)
            DFS(0, i);

        for(int i=0; i <n; i++)
        {
            ans[i] = dp[0][i];
            if(a[i] == b[i])
                ans[i] =i?ans[i-1]:0;


            for(int j=0; j<i; j++)
                ans[i] = min(ans[i], ans[j]+dp[j+1][i]);
        }
        printf("%d\n", ans[n-1]);
    }
    return 0;
}

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const LL INF = 0xfffffff;
const LL maxn = 105;
int dp[maxn][maxn], ans[maxn];
char a[maxn], b[maxn];

int main()
{

    while(cin >> a >> b)
    {
        memset(dp, 0, sizeof(dp));
        int n = strlen(a);        for(int len=0; len<n; len++)
        {
            for(int i=0; i+len<n; i++)
            {
                int j = i + len;
                dp[i][j] = dp[i+1][j] + 1;
                for(int k=i+1; k<=j; k++)
                {
                    if(b[i] == b[k])
                    {
                        dp[i][j] = min(dp[i][j], dp[i+1][k] + dp[k+1][j]);
                    }
                }
            }
        }

        for(int i=0; i <n; i++)
        {
            ans[i] = dp[0][i];
            if(a[i] == b[i])
                ans[i] =i?ans[i-1]:0;


            for(int j=0; j<i; j++)
                ans[i] = min(ans[i], ans[j]+dp[j+1][i]);
        }
        printf("%d\n", ans[n-1]);
    }
    return 0;
}