uva 10453 Make Palindrome （区间DP + 递归输出）

uva 10453 Make Palindrome

题目大意：给出一段字符串，要求求出最少加入几个字符（任意位置），可以让该字符串变成会问字符串，并输出修改以后的回文字符串。

解题思路：dp[i][j]代表了将该字符串从第i位到第j位变成回文字符串最少要添加的字符。当S[i]==S[j],dp[i][j]=dp[i+1][j−1]S[i] == S[j], dp[i][j] = dp[i + 1][ j - 1]当S[i]!=S[j],dp[i][j]=min(dp[i+1][j],dp[i][j−1])+1S[i] != S[j] ,dp[i][j] = min(dp[i + 1][j], dp[i][j - 1]) + 1,在DP的过程中记录对该区间的操作类型，最后递归输出。

[code]#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#define N 1005
using namespace std;
typedef long long ll;
char s
;
int dp

, rec

, len;
void DP() {
    len = strlen(s);
    memset(dp, 0, sizeof(dp));
    for (int i = len - 1; 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 {
                if (dp[i + 1][j] < dp[i][j - 1]) {
                    dp[i][j] = dp[i + 1][j] + 1;
                    rec[i][j] = -1;
                } else {
                    dp[i][j] = dp[i][j - 1] + 1;
                    rec[i][j] = 1;
                }
            }
        }
    }
}
void myprint(int a, int b) {
    if (a > b) return;
    if (a == b) printf("%c", s[a]);
    else if (rec[a][b] == 0) {
        printf("%c", s[a]);
        myprint(a + 1, b - 1);
        printf("%c", s[a]);
    } else if (rec[a][b] == 1) {
        printf("%c", s[b]);
        myprint(a, b - 1);
        printf("%c", s[b]);
    } else if (rec[a][b] == -1) {
        printf("%c", s[a]);
        myprint(a + 1, b);
        printf("%c", s[a]);
    }
}
int main() {
    while (scanf("%s", s) != EOF) {
        memset(rec, 0, sizeof(rec));
        DP();
        printf("%d ", dp[0][len - 1]);
        myprint(0, len - 1);
        printf("\n");
    }
    return 0;
}