[hdu4628 Pieces]二进制子状态，DP

题意：给一个长度为16的字符串，每次从里面删掉一个回文序列，求最少需要几次才能删掉所有字符

思路：二进制表示每个字符的状态，那么从1个状态到另一个状态有两种转移方式，一是枚举所有合法的回文子序列，判断是否是当前状态的子状态，再转移，二是枚举当前状态的所有子状态来转移。前者最坏复杂度O(2^16*2^16) = O(几十亿)，而后者最坏只有(i:1->16)Σ2iC(16,i) = O(几千万）。

 

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#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
template<typename T>
void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}
template<typename T>
void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;

/* -------------------------------------------------------------------------------- */

char s[20];
int n, dp[1 << 20];

void init() {
    for (int i = 0; i < (1 << n); i ++) dp[i] = INF;
    dp[0] = 1;
    for (int i = 1; i < (1 << n); i ++) {
        int j, k;
        for (j = n - 1; j >= 0; j --) {
            if (i & (1 << j)) break;
        }
        for (k = 0; k < n; k ++) {
            if (i & (1 << k)) break;
        }
        if (s[j] == s[k] && dp[i ^ (1 << j) ^ (1 << k)] < INF)
            dp[i] = dp[i ^ (1 << j) ^ (1 << k)];
        if (j == k) dp[i] = 1;
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int T;
    cin >> T;
    while (T --) {
        scanf("%s", s);
        n = strlen(s);
        init();
        for (int i = 1; i < (1 << n); i ++) {
            for (int j = i; j; j = (j - 1) & i) {
                umin(dp[i], dp[i ^ j] + dp[j]);
            }
        }
        cout << dp[(1 << n) - 1] << endl;
    }
    return 0;
}
posted @ 2015-08-12 15:21  jklongint  阅读(265)  评论(0)  编辑  收藏  举报