UVA 10453 - Make Palindrome

来源：互联网发布：用友nc6.5数据库编辑：程序博客网时间：2024/06/05 03:48

这个题也是很简单的dp了。

状态转移方程就比较简单了。

在输出的时候也是递归写的。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <algorithm>#include <fstream>#include <set>#include <map>#include <queue>#include <stack>#include <list>#include <vector>#include <cmath>#include <iomanip>typedef long long LL;typedef unsigned long long LLU;const double PI=acos(-1.0);using namespace std;#define MAXN 1000+10int dp[MAXN][MAXN] = {0};char s[MAXN];void dfs(int i, int j){    if(i > j)        return ;    else if(i == j){        printf("%c",s[i]);        return ;    }    if(s[i] == s[j]){        printf("%c",s[i]);        dfs(i+1,j-1);        printf("%c",s[i]);    }    else{        if(dp[i+1][j] < dp[i][j-1]){            printf("%c",s[i]);            dfs(i+1,j);            printf("%c",s[i]);        }        else{            printf("%c",s[j]);            dfs(i,j-1);            printf("%c",s[j]);        }    }}int main (){    while(gets(s) != NULL){        int len = strlen(s);        memset(dp,0,sizeof(dp));        for(int i = len-2; i >= 0; i--){            for(int j = i+1; j < len; j++){                if(s[i] == s[j])                    dp[i][j] = dp[i+1][j-1];                else                    dp[i][j] = min(dp[i+1][j],dp[i][j-1]) + 1;            }        }        printf("%d ",dp[0][len-1]);        dfs(0,len-1);        printf("\n");    }    return 0;}

0 0