D. Minimax Problem
time limit per test
5 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output
You are given nn arrays a1a1, a2a2, ..., anan; each array consists of exactly mm integers. We denote the yy-th element of the xx-th array as ax,yax,y.
You have to choose two arrays aiai and ajaj (1≤i,j≤n1≤i,j≤n, it is possible that i=ji=j). After that, you will obtain a new array bb consisting of mmintegers, such that for every k∈[1,m]k∈[1,m] bk=max(ai,k,aj,k)bk=max(ai,k,aj,k).
Your goal is to choose ii and jj so that the value of mink=1mbkmink=1mbk is maximum possible.
Input
The first line contains two integers nn and mm (1≤n≤3⋅1051≤n≤3⋅105, 1≤m≤81≤m≤8) — the number of arrays and the number of elements in each array, respectively.
Then nn lines follow, the xx-th line contains the array axax represented by mm integers ax,1ax,1, ax,2ax,2, ..., ax,max,m (0≤ax,y≤1090≤ax,y≤109).
Output
Print two integers ii and jj (1≤i,j≤n1≤i,j≤n, it is possible that i=ji=j) — the indices of the two arrays you have to choose so that the value of mink=1mbkmink=1mbk is maximum possible. If there are multiple answers, print any of them.
Example
input
Copy
6 5 5 0 3 1 2 1 8 9 1 3 1 2 3 4 5 9 1 0 3 7 2 3 0 6 3 6 4 1 7 0
output
Copy
1 5
1 /*
2 看到题目首先想到的就是二分答案，那么找到一个可能的答案之后就可以把所有数列转化成01数列，1代表该位置的数大于等于答案，0代表该位置的数小于答案。
3
4 选择两个数列使其满足答案，也就是使这两个数列的或和为（1<<m）-1，很显然不能n*n的枚举每个数列，但是2^8只有256，所以可以直接枚举数列对应的二进制数，判断答案是否可行，如果可行，就可以O（n）地寻找可行解
5 */
6 #include<iostream>
7 #include<cstdio>
8 #include<bitset>
9 #include<cstring>
10 using namespace std;
11 int n,m,a[300010][10],ans1,ans2;
12 int b[300010];
13 bool vis[300];
14 bool check(int x){
15 memset(b,0,sizeof(b));
16 for(int i=1;i<=n;i++){
17 for(int j=1;j<=m;j++){
18 if(a[i][j]>=x)b[i]^=1<<(j-1);
19 }
20 }
21 memset(vis,0,sizeof(vis));
22 for(int i=1;i<=n;i++)vis[b[i]]=1;
23 bool flag=0;
24 int s1=0,s2=0;
25 for(int sta1=0;sta1<(1<<m);sta1++){
26 for(int sta2=sta1;sta2<(1<<m);sta2++){
27 if((sta1|sta2)==(1<<m)-1&&vis[sta1]&&vis[sta2]){
28 s1=sta1;s2=sta2;
29 flag=1;
30 break;
31 }
32 }
33 if(flag)break;
34 }
35 if(!flag)return 0;
36 for(int i=1;i<=n;i++){
37 if(b[i]==s1){
38 ans1=i;
39 break;
40 }
41 }
42 for(int i=1;i<=n;i++){
43 if(b[i]==s2){
44 ans2=i;
45 break;
46 }
47 }
48 return 1;
49 }
50 int main(){
51 scanf("%d%d",&n,&m);
52 for(int i=1;i<=n;i++)
53 for(int j=1;j<=m;j++){
54 scanf("%d",&a[i][j]);
55 }
56 int l=0,r=1000000000;
57 while(l<=r){
58 int mid=(l+r)>>1;
59 if(check(mid)){
60 l=mid+1;
61 }
62 else r=mid-1;
63 }
64 printf("%d %d\n",ans1,ans2);
65 return 0;
66 }
原文地址：https://www.cnblogs.com/thmyl/p/12237546.html