hdu5323 Solve this interesting problem(爆搜)

转载请注明出处： http://www.cnblogs.com/fraud/ ——by fraud

Solve this interesting problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1731 Accepted Submission(s): 519

Problem Description

Have you learned something about segment tree? If not, don’t worry, I will explain it for you.
Segment Tree is a kind of binary tree, it can be defined as this:
- For each node u in Segment Tree, u has two values:

Input

The input consists of several test cases.
Each test case contains two integers L and R, as described above.

Output

For each test, output one line contains one integer. If there is no such n, just output -1.

Sample Input

6 7 10 13 10 11

Sample Output

7 -1 12

比较隐晦的指出了爆搜的最大深度一定不会超过11层，所以，复杂度就是4^11次方，然后加一点可行性的剪枝之类，就能AC了

 1 //#####################
 2 //Author:fraud
 3 //Blog: http://www.cnblogs.com/fraud/
 4 //#####################
 5 //#pragma comment(linker, "/STACK:102400000,102400000")
 6 #include <iostream>
 7 #include <sstream>
 8 #include <ios>
 9 #include <iomanip>
10 #include <functional>
11 #include <algorithm>
12 #include <vector>
13 #include <string>
14 #include <list>
15 #include <queue>
16 #include <deque>
17 #include <stack>
18 #include <set>
19 #include <map>
20 #include <cstdio>
21 #include <cstdlib>
22 #include <cmath>
23 #include <cstring>
24 #include <climits>
25 #include <cctype>
26 using namespace std;
27 #define XINF INT_MAX
28 #define INF 0x3FFFFFFF
29 #define MP(X,Y) make_pair(X,Y)
30 #define PB(X) push_back(X)
31 #define REP(X,N) for(int X=0;X<N;X++)
32 #define REP2(X,L,R) for(int X=L;X<=R;X++)
33 #define DEP(X,R,L) for(int X=R;X>=L;X--)
34 #define CLR(A,X) memset(A,X,sizeof(A))
35 #define IT iterator
36 typedef long long ll;
37 typedef pair<int,int> PII;
38 typedef vector<PII> VII;
39 typedef vector<int> VI;
40 ll ans = 0;
41 void dfs(ll l,ll r){
42     if(l<0)return;
43     if(r<l)return;
44     if(l==0){
45         if(ans==-1)ans = r;
46         else ans = min(ans,r);
47         return;
48     }
49     if(r>=ans&&ans!=-1)return;
50     if((r-l+1)>(l))return;
51     dfs(l-(r-l)-2,r);
52     dfs(l,r+(r-l));
53     dfs(l-(r-l)-1,r);
54     dfs(l,r+(r-l)+1);
55     return;
56 }
57 int main()
58 {
59     ios::sync_with_stdio(false);
60     ll l,r;
61     while(cin>>l>>r){
62         ans = -1;
63         dfs(l,r);
64         cout<<ans<<endl;
65     }
66     return 0;
67 }

posted on 2015-07-29 19:43 xyiyy 阅读(243) 评论(0) 编辑收藏举报

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