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