[Uva-11584] partitioning by palindromes (DP)

The rough complexity is L ^ 3, and the maximum length is 1000. I did not dare to do it. Later, I found that the complexity coefficient is not big, and it can pass through very quickly.

DP [J] = DP [I-1] + 1 (if (STR [I] ~ STR [J] is a reply)

14327451

11584

Partitioning by palindromes

Accepted

C ++

0.052

09:33:17

#include<set>#include<map>#include<stack>#include<queue>#include<vector>#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>using namespace std;typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> pill;const int maxn = 1111;char str[maxn];int   dp[maxn];bool is_part(int i,int j){    while(i < j){        if(str[i] != str[j])            return false;        i ++;        j --;    }    return true;}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%s",str);        int L = strlen(str);        for(int i = 0 ; i <= L; i ++) dp[i] = i + 1;        for(int j = 0 ; j < L; j ++) // L            for(int i = 0 ; i <= j ; i ++){ // L                if(is_part(i,j)){ // L                    if(i - 1 >= 0)                        dp[j] = min(dp[j],dp[i - 1] + 1);                    else                        dp[j] = 1;                }            }        printf("%d\n",dp[L - 1]);    }    return 0;}

[Uva-11584] partitioning by palindromes (DP)