URAL 1635. Mnemonics and Palindromes(DP)

题目链接

本来用区间DP，3次方的复杂度，T了，看了看题解，降维，直接二次方的复杂度可以解。然后折腾一下输出路径。。终于过了。

#include <cstring>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
char str[4001];
int dp[4001][4001];
int p[4001];
int pre[4001];
int flag[4001];
int main()
{
    int i,len,j;
    scanf("%s",str);
    len = strlen(str);
    for(i = 0; i <= len; i ++)
    {
        dp[i][i] = 1;
        for(j = 1;; j ++)
        {
            if(i-j+1 < 0)break;
            if(i+j >= len) break;
            if(str[i-j] != str[i+j]) break;
            dp[i-j][i+j] = 1;
        }
        for(j = 1;; j ++)
        {
            if(i-j+1 < 0)break;
            if(i+j >= len) break;
            if(str[i-j+1] != str[i+j]) break;
            dp[i-j+1][i+j] = 1;
        }
    }
    for(i = 0; i < len; i ++)
        p[i] = 100000000;
    p[0] = 1;
    for(i = 1; i < len; i ++)
    {
        if(dp[0][i])
        {
            pre[i] = 0;
            p[i] = 1;
        }
    }
    for(i = 0; i < len; i ++)
    {
        if(p[i] == 1) continue;
        for(j = 0; j < i; j ++)
        {
            if(dp[j+1][i])
            {
                if(p[i] > p[j] + 1)
                {
                    pre[i] = j+1;
                    p[i] = p[j]+1;
                }
            }
        }
    }
    printf("%d
",p[len-1]);
    int a = pre[len-1];
    if(a != 0)
    flag[a] = 1;
    a --;
    while(p[len-1] >= 2)
    {
        a = pre[a];
        if(a != 0)
        flag[a] = 1;
        a --;
        p[len-1] --;
    }
    for(i = 0; i < len; i ++)
    {
        if(flag[i]) printf(" ");
        printf("%c",str[i]);
    }
    printf("
");
    return 0;
}