HDU 2476 String painter (区间DP)

题意：给定两个串，问你从第一个串刷成第二个串最少要几次。

析：我们可以先求一个空串变成第二个串，然后再求第一个串的，dp[i][j] 表示 i-j 这个区间已经和第二个串相同了，最少要几次，区间dp么，

然后再求和第一个的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn][maxn], ans[maxn];
char a[maxn], b[maxn];

int main(){
  while(scanf("%s", a) == 1){
    scanf("%s", b);
    memset(dp, 0, sizeof dp);
    n = strlen(a);
    for(int i = 0; i < n; ++i)  dp[i][i] = 1;
    for(int l = 1; l < n; ++l)
      for(int i = 0; i + l < n; ++i){
        int j = i + l;
        dp[i][j] = dp[i+1][j] + 1;
        for(int k = i+1; k <= j; ++k)
          dp[i][j] = min(dp[i][j], dp[i+1][k] + dp[k+1][j] + (b[i] != b[k]));
      }

    for(int i = 0; i < n; ++i){
      ans[i] = dp[0][i];
      if(a[i] == b[i])  ans[i] = i == 0 ? 0 : ans[i-1];
      for(int j = 0; j < i; ++j)
        ans[i] = min(ans[i], ans[j] + dp[j+1][i]);
    }
    printf("%d
", ans[n-1]);
  }
  return 0;
}