SRM 584 div2

早早地水完了三道题，pt1000用的是dfs，开始做的时候误认为复杂度最多就O(2^25)，结果被一组O(2*3^16)的数据接近1e8给cha了。继续努力。

pt250：求两个串的前缀组成的不同串数目。set搞定。

/* 
 *Author:       Zhaofa Fang 
 *Created time: 2013-07-10-18.54 
 *Language:     C++ 
 */ 
#include <cstdio> 
#include <cstdlib> 
#include <sstream> 
#include <iostream> 
#include <cmath> 
#include <cstring> 
#include <algorithm> 
#include <string> 
#include <utility> 
#include <vector> 
#include <queue> 
#include <map> 
#include <set> 
using namespace std; 

typedef long long ll; 
#define DEBUG(x) cout<< #x << ':' << x << endl 
#define FOR(i,s,t) for(int i = (s);i <= (t);i++) 
#define FORD(i,s,t) for(int i = (s);i >= (t);i--) 
#define REP(i,n) for(int i=0;i<(n);i++) 
#define REPD(i,n) for(int i=(n-1);i>=0;i--) 
#define PII pair<int,int> 
#define PB push_back 
#define MP make_pair 
#define ft first 
#define sd second 
#define lowbit(x) (x&(-x)) 
#define INF (1<<30) 
#define eps 1e-8 

class TopFox{ 
public : 
    int possibleHandles(string f, string g){ 
        set<string>S; 
        int lenf = f.length(); 
        int leng = g.length(); 
        REP(i,lenf){ 
            REP(j,leng){ 
                string tmp = ""; 
                REP(k,i+1)tmp += f[k]; 
                REP(k,j+1)tmp += g[k]; 
                S.insert(tmp); 
            } 
        } 
        return S.size(); 
    } 
}; 

pt500：给定一幅图的连接情况，要求相邻两个点的差不大于d，求点之间的最大差。

　　　　可以对每个点进行bfs，比较直观。也有一些人用的是floyd，道理一样的。

/* 
 *Author:       Zhaofa Fang 
 *Created time: 2013-07-10-18.54 
 *Language:     C++ 
 */ 
#include <cstdio> 
#include <cstdlib> 
#include <sstream> 
#include <iostream> 
#include <cmath> 
#include <cstring> 
#include <algorithm> 
#include <string> 
#include <utility> 
#include <vector> 
#include <queue> 
#include <map> 
#include <set> 
using namespace std; 

typedef long long ll; 
#define DEBUG(x) cout<< #x << ':' << x << endl 
#define FOR(i,s,t) for(int i = (s);i <= (t);i++) 
#define FORD(i,s,t) for(int i = (s);i >= (t);i--) 
#define REP(i,n) for(int i=0;i<(n);i++) 
#define REPD(i,n) for(int i=(n-1);i>=0;i--) 
#define PII pair<int,int> 
#define PB push_back 
#define MP make_pair 
#define ft first 
#define sd second 
#define lowbit(x) (x&(-x)) 
#define INF (1<<30) 
#define eps 1e-8 

class Egalitarianism{ 
public : 
    bool vist[100]; 
    int mon[100]; 
    bool OK; 
    int bfs(int s,vector <string> isFriend, int d){ 
        int n = isFriend.size(); 
        queue<int>Q; 
        Q.push(s); 
        int ans = -1; 
        memset(vist,0,sizeof(vist)); 
        mon[s] = 0; 
        vist[s] = 1; 
        while(!Q.empty()){ 
            int now = Q.front(); 
            Q.pop(); 
            REP(i,n){ 
                if(isFriend[now][i] == 'Y' && !vist[i]){ 
                    Q.push(i); 
                    mon[i] = mon[now] + d; 
                    vist[i] = 1; 
                    ans = max(ans,mon[i]); 
                } 
            } 
        } 
        REP(i,n)if(!vist[i])return -1; 
        return ans; 
    } 
    int maxDifference(vector <string> isFriend, int d){ 
        int n = isFriend.size(); 
        int ans = -1; 
        REP(i,n)ans = max(bfs(i,isFriend,d),ans); 
        return ans; 
    } 
}; 

pt1000：给定一组数kind[],从中取K个，取得的数为found[]，能有多少种可能。其中一组sample如下：

{1, 2, 1, 1, 2, 3, 4, 3, 2, 2}

{4, 2}

3

Returns: 6

The archeologist excavated one of six possible triples of building sites:

• (1, 4, 6)
• (1, 6, 8)
• (1, 6, 9)
• (4, 6, 8)
• (4, 6, 9)
• (6, 8, 9)

　　正解dp。很简单的转移-__-。

/*
 *Author:       Zhaofa Fang
 *Created time: 2013-07-10-18.54
 *Language:     C++
 */
#include <cstdio>
#include <cstdlib>
#include <sstream>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <utility>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std;

typedef long long ll;
#define DEBUG(x) cout<< #x << ':' << x << endl
#define FOR(i,s,t) for(int i = (s);i <= (t);i++)
#define FORD(i,s,t) for(int i = (s);i >= (t);i--)
#define REP(i,n) for(int i=0;i<(n);i++)
#define REPD(i,n) for(int i=(n-1);i>=0;i--)
#define PII pair<int,int>
#define PB push_back
#define MP make_pair
#define ft first
#define sd second
#define lowbit(x) (x&(-x))
#define INF (1<<30)
#define eps 1e-8

ll C[55][55];

void pre(){
    C[1][0] = C[1][1] = 1;
    FOR(i,2,50){
        C[i][0] = 1;
        FOR(j,1,i)C[i][j] = C[i-1][j-1] + C[i-1][j];
    }
}
class Excavations2{
public :
    int ff[55];
    ll dp[55][55];
    ll count(vector <int> kind, vector <int> found, int K){
        pre();
        int len = kind.size();
        memset(ff,0,sizeof(ff));
        REP(i,len)ff[kind[i]]++;
        memset(dp,0,sizeof(dp));
        int n = found.size();
        FOR(i,1,ff[found[0]]){
            dp[1][i] = C[ff[found[0]]][i];
        }
        FOR(i,2,n){
            FOR(j,1,K){
                FOR(k,1,ff[found[i-1]])
                    dp[i][j] += dp[i-1][j-k]*C[ff[found[i-1]]][k];
            }
        }
        return dp[n][K];
    }
};


//int main(){
//    //freopen("in","r",stdin);
//    //freopen("out","w",stdout);
//
//    return 0;
//}