[Ultraviolet A] 10739-string to palindrome (Dynamic Planning)

Dynamic Comparison Planning

DP [I] [J] convert the original string I ~ Minimum number of operations required to convert a character within J to a return character

The Delete and add operations are essentially the same.

Three state transition equations:

DP [I] [J] = min (DP [I] [J], DP [I + 1] [J]);

DP [I] [J] = min (DP [I] [J], DP [I + 1] [J-1]);

DP [I] [J] = min (DP [I] [J], DP [I] [J-1]);

If I = j dp [I] [J] = 0;

14145138

10651

Pebble Solitaire

Accepted

C ++

0.009

2014-09-04 09:09:42

#include<cstdio>#include<algorithm>#include<string>#include<cstring>#include<map>#include<iostream>using namespace std;#define MAXD 1000 + 10#define INF 10000char str[MAXD];int dp[MAXD][MAXD];int dfs(int start,int last){    if(dp[start][last] != -1)        return dp[start][last];    if(start == last)        return dp[start][last] = 0;    if(str[start] == str[last]){        if(start + 1 == last)            return dp[start][last] = 0;        else            return dp[start][last] = dfs(start + 1 , last - 1);    }    dp[start][last] = INF;    if(last - 1 >= start)    dp[start][last] = min(dp[start][last],dfs(start,last - 1) + 1);    if(start + 1 <= last)    dp[start][last] = min(dp[start][last],dfs(start + 1, last) + 1);    if(start + 1 <= last - 1)    dp[start][last] = min(dp[start][last],dfs(start + 1,last - 1) + 1);    return dp[start][last];}int main(){    int T;    scanf("%d",&T);    for(int Case = 1; Case <= T; Case ++){        scanf("%s",str);        memset(dp,-1,sizeof(dp));        int ans = dfs(0,strlen(str) - 1);        printf("Case %d: %d\n",Case,ans);    }    return 0;}

[Ultraviolet A] 10739-string to palindrome (Dynamic Planning)