HDU 4628 Pieces（状压DP）题解

Posted 2021-12-16 kirinsb

题意：n个字母，每次可以删掉一组非连续回文，问你最少删几次

思路：把所有回文找出来，然后状压DP

代码：

#include<set>
#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include <iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 16 + 5;
const int M = maxn * 30;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
const int MOD = 1e4 + 7;
char s[maxn];
int pa[1 << maxn];
int dp[1 << maxn];
int n, cnt;
bool check(int st)
    char tmp[20];
    int len = 0;
    for(int i = 0; i < n; i++)
        if(st & (1 << i))
            tmp[++len] = s[i];
        
    
    for(int i = 1; i <= len / 2; i++)
        if(tmp[i] != tmp[len - i + 1]) return false;
    
    return true;

int main()
    int T;
    scanf("%d", &T);
    while(T--)
        scanf("%s", &s);
        n = strlen(s);
        cnt = 0;
        memset(dp, INF, sizeof(dp));
        for(int i = 1; i < (1 << n); i++)
            if(check(i))
                pa[cnt++] = i;
                dp[i] = 1;
            
        
        for(int i = 0; i < (1 << n); i++)
            for(int j = 0; j < cnt; j++)
                if(i & pa[j]) continue;
                dp[i | pa[j]] = min(dp[i | pa[j]], dp[i] + 1);
            
        
        printf("%d\n", dp[(1 << n) - 1]);
    
    return 0;