CF1288D Minimax Problem（二分 + 状态压缩）

Minimax Problem

思路：我们通过二分去试探答案，假设答案是x，则我们把数组中大于等于key的标记为1，反之标为0。我们知道m = 8，则01串的情况只有2^8-1种，那么我们可以把n行的情况压缩在vis[2^8-1]长度的数组中，vis记录符合的下标，如果（!vis[x] && !vis[y] && (x | y) == (2^8 -1)）说明这个key是符合的，我们继续通过二分去试探更大的答案。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <functional>
 5 #include <set>
 6 #include <vector>
 7 #include <queue>
 8 #include <cstring>
 9 #include <stack>
10  
11 using namespace std;
12  
13 #define ll long long
14 #define pb push_back
15 #define fi first
16 #define se second
17  
18 vector<vector<int > > a;
19 vector<int > vis;
20 int n, m;
21 int fi, se;
22 
23 bool check(int mid){
24     fill(vis.begin(), vis.end(), 0);
25     for(int i = 1; i <= n; ++i){
26 
27         int cur = 0;
28         for(int j = 1; j <= m; ++j){
29             if(a[i][j] >= mid) cur |= (1 << (j - 1));
30         }
31         //cout << "cur = " << cur << endl;
32         vis[cur] = i;
33     }
34 
35     for(int i = 0; i < (1 << m); ++i){
36         for(int j = 0; j < (1 << m); ++j){
37             if(vis[i] && vis[j] && (i | j) == (1 << m) - 1){
38                 fi = vis[i];
39                 se = vis[j];
40                 return true;
41             }
42         }
43     }
44     return false;
45 }
46 
47 void solve(){
48     cin >> n >> m;
49     a.resize(n + 1, vector<int >(m + 1));
50     vis.resize((1 << m) + 10);
51     for(int i = 1; i <= n; ++i){
52         for(int j = 1; j <= m; ++j)
53             cin >> a[i][j];
54     }
55     int l = 0, r = (int)1e9 + 10;
56     while(l <= r){
57         int mid = (l + r) >> 1;
58         //cout << mid << endl;
59         if(check(mid)) l = mid + 1;
60         else r = mid - 1;
61     }
62     cout << fi << " " << se << endl; 
63 }
64  
65 int main(){
66     
67     // freopen("C:\\Users\\admin\\Desktop\\input.txt", "r", stdin);
68     // freopen("C:\\Users\\admin\\Desktop\\output.txt", "w", stdout);
69     ios::sync_with_stdio(false);
70     cin.tie(0);
71     cout.tie(0);
72     solve();
73     
74     return 0;
75 }

 

posted @ 2020-05-05 12:22  SummerMingQAQ  阅读(230)  评论(0)  编辑  收藏  举报