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+)tmp += f[k];
                 REP(k,j+)tmp += g[k];
                 S.insert(tmp);
             }
         }
         return S.size();
     }
 }; 

250

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[];
     int mon[];
     bool OK;
     int bfs(int s,vector <string> isFriend, int d){
         int n = isFriend.size();
         queue<int>Q;
         Q.push(s);
         int ans = -;
         memset(vist,,sizeof(vist));
         mon[s] = ;
         vist[s] = ;
         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] = ;
                     ans = max(ans,mon[i]);
                 }
             }
         }
         REP(i,n)if(!vist[i])return -;
         return ans;
     }
     int maxDifference(vector <string> isFriend, int d){
         int n = isFriend.size();
         int ans = -;
         REP(i,n)ans = max(bfs(i,isFriend,d),ans);
         return ans;
     }
 }; 

500

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[][];

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

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

1000