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。

 1 /* 
 2  *Author:       Zhaofa Fang 
 3  *Created time: 2013-07-10-18.54 
 4  *Language:     C++ 
 5  */ 
 6 #include <cstdio> 
 7 #include <cstdlib> 
 8 #include <sstream> 
 9 #include <iostream> 
10 #include <cmath> 
11 #include <cstring> 
12 #include <algorithm> 
13 #include <string> 
14 #include <utility> 
15 #include <vector> 
16 #include <queue> 
17 #include <map> 
18 #include <set> 
19 using namespace std; 
20 
21 typedef long long ll; 
22 #define DEBUG(x) cout<< #x << ':' << x << endl 
23 #define FOR(i,s,t) for(int i = (s);i <= (t);i++) 
24 #define FORD(i,s,t) for(int i = (s);i >= (t);i--) 
25 #define REP(i,n) for(int i=0;i<(n);i++) 
26 #define REPD(i,n) for(int i=(n-1);i>=0;i--) 
27 #define PII pair<int,int> 
28 #define PB push_back 
29 #define MP make_pair 
30 #define ft first 
31 #define sd second 
32 #define lowbit(x) (x&(-x)) 
33 #define INF (1<<30) 
34 #define eps 1e-8 
35 
36 class TopFox{ 
37 public : 
38     int possibleHandles(string f, string g){ 
39         set<string>S; 
40         int lenf = f.length(); 
41         int leng = g.length(); 
42         REP(i,lenf){ 
43             REP(j,leng){ 
44                 string tmp = ""; 
45                 REP(k,i+1)tmp += f[k]; 
46                 REP(k,j+1)tmp += g[k]; 
47                 S.insert(tmp); 
48             } 
49         } 
50         return S.size(); 
51     } 
52 };
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。

 1 /* 
 2  *Author:       Zhaofa Fang 
 3  *Created time: 2013-07-10-18.54 
 4  *Language:     C++ 
 5  */ 
 6 #include <cstdio> 
 7 #include <cstdlib> 
 8 #include <sstream> 
 9 #include <iostream> 
10 #include <cmath> 
11 #include <cstring> 
12 #include <algorithm> 
13 #include <string> 
14 #include <utility> 
15 #include <vector> 
16 #include <queue> 
17 #include <map> 
18 #include <set> 
19 using namespace std; 
20 
21 typedef long long ll; 
22 #define DEBUG(x) cout<< #x << ':' << x << endl 
23 #define FOR(i,s,t) for(int i = (s);i <= (t);i++) 
24 #define FORD(i,s,t) for(int i = (s);i >= (t);i--) 
25 #define REP(i,n) for(int i=0;i<(n);i++) 
26 #define REPD(i,n) for(int i=(n-1);i>=0;i--) 
27 #define PII pair<int,int> 
28 #define PB push_back 
29 #define MP make_pair 
30 #define ft first 
31 #define sd second 
32 #define lowbit(x) (x&(-x)) 
33 #define INF (1<<30) 
34 #define eps 1e-8 
35 
36 class Egalitarianism{ 
37 public : 
38     bool vist[100]; 
39     int mon[100]; 
40     bool OK; 
41     int bfs(int s,vector <string> isFriend, int d){ 
42         int n = isFriend.size(); 
43         queue<int>Q; 
44         Q.push(s); 
45         int ans = -1; 
46         memset(vist,0,sizeof(vist)); 
47         mon[s] = 0; 
48         vist[s] = 1; 
49         while(!Q.empty()){ 
50             int now = Q.front(); 
51             Q.pop(); 
52             REP(i,n){ 
53                 if(isFriend[now][i] == 'Y' && !vist[i]){ 
54                     Q.push(i); 
55                     mon[i] = mon[now] + d; 
56                     vist[i] = 1; 
57                     ans = max(ans,mon[i]); 
58                 } 
59             } 
60         } 
61         REP(i,n)if(!vist[i])return -1; 
62         return ans; 
63     } 
64     int maxDifference(vector <string> isFriend, int d){ 
65         int n = isFriend.size(); 
66         int ans = -1; 
67         REP(i,n)ans = max(bfs(i,isFriend,d),ans); 
68         return ans; 
69     } 
70 };
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-__-。

 1 /*
 2  *Author:       Zhaofa Fang
 3  *Created time: 2013-07-10-18.54
 4  *Language:     C++
 5  */
 6 #include <cstdio>
 7 #include <cstdlib>
 8 #include <sstream>
 9 #include <iostream>
10 #include <cmath>
11 #include <cstring>
12 #include <algorithm>
13 #include <string>
14 #include <utility>
15 #include <vector>
16 #include <queue>
17 #include <map>
18 #include <set>
19 using namespace std;
20 
21 typedef long long ll;
22 #define DEBUG(x) cout<< #x << ':' << x << endl
23 #define FOR(i,s,t) for(int i = (s);i <= (t);i++)
24 #define FORD(i,s,t) for(int i = (s);i >= (t);i--)
25 #define REP(i,n) for(int i=0;i<(n);i++)
26 #define REPD(i,n) for(int i=(n-1);i>=0;i--)
27 #define PII pair<int,int>
28 #define PB push_back
29 #define MP make_pair
30 #define ft first
31 #define sd second
32 #define lowbit(x) (x&(-x))
33 #define INF (1<<30)
34 #define eps 1e-8
35 
36 ll C[55][55];
37 
38 void pre(){
39     C[1][0] = C[1][1] = 1;
40     FOR(i,2,50){
41         C[i][0] = 1;
42         FOR(j,1,i)C[i][j] = C[i-1][j-1] + C[i-1][j];
43     }
44 }
45 class Excavations2{
46 public :
47     int ff[55];
48     ll dp[55][55];
49     ll count(vector <int> kind, vector <int> found, int K){
50         pre();
51         int len = kind.size();
52         memset(ff,0,sizeof(ff));
53         REP(i,len)ff[kind[i]]++;
54         memset(dp,0,sizeof(dp));
55         int n = found.size();
56         FOR(i,1,ff[found[0]]){
57             dp[1][i] = C[ff[found[0]]][i];
58         }
59         FOR(i,2,n){
60             FOR(j,1,K){
61                 FOR(k,1,ff[found[i-1]])
62                     dp[i][j] += dp[i-1][j-k]*C[ff[found[i-1]]][k];
63             }
64         }
65         return dp[n][K];
66     }
67 };
68 
69 
70 //int main(){
71 //    //freopen("in","r",stdin);
72 //    //freopen("out","w",stdout);
73 //
74 //    return 0;
75 //}
1000

 

 

posted @ 2013-07-11 12:42  發_  阅读(274)  评论(0)  编辑  收藏  举报