Codeforces1102F Elongated Matrix 状压DP

Posted 2021-02-04 menhera

题目分析：

这题瞎搞一个哈密尔顿路，对于起点不同的分开跑就可以过了。

$O(n^3*2^n)$

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn = 17;
 5 const int maxm = 10200;
 6 
 7 int n,m;
 8 int a[maxn][maxm],f[maxn][maxn],g[maxn][maxn];
 9 
10 int dp[maxn][(1<<16)+5],arr[maxn][(1<<16)+5];
11 
12 void init(){
13     for(int i=1;i<=n;i++){
14     for(int j=1;j<=n;j++){
15         f[i][j] = g[i][j] = 1e9;
16         for(int k=1;k<=m;k++)f[i][j] = min(f[i][j],abs(a[i][k]-a[j][k]));
17         for(int k=1;k<m;k++) g[i][j]=min(g[i][j],abs(a[i][k+1]-a[j][k]));
18     }
19     }
20 }
21 
22 queue<pair<int,int> > q;
23 void divide(){
24     int ans = 0;
25     for(int i=1;i<=n;i++){
26     memset(dp,0,sizeof(dp));
27     memset(arr,0,sizeof(arr));
28     dp[i][1<<i-1] = 1e9; arr[i][1<<i-1] = 1;
29     q.push(make_pair(i,1<<i-1));
30     while(!q.empty()){
31         pair<int,int> k = q.front(); q.pop();
32         for(int j=1;j<=n;j++){
33         if(k.second&(1<<j-1)) continue;
34         dp[j][k.second|(1<<j-1)] =
35             max(dp[j][k.second|(1<<j-1)],min(dp[k.first][k.second],f[k.first][j]));
36         if(!arr[j][k.second|(1<<j-1)]){
37             q.push(make_pair(j,k.second|(1<<j-1)));
38             arr[j][k.second|(1<<j-1)] = 1;
39         }
40         }
41     }
42     for(int j=1;j<=n;j++){
43         if(i == j) continue;
44         ans = max(ans,min(dp[j][(1<<n)-1],g[i][j]));
45     }
46     }
47     printf("%d
",ans);
48 }
49 
50 void read(){
51     scanf("%d%d",&n,&m);
52     for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%d",&a[i][j]);
53 }
54 
55 int main(){
56     read();
57     if(n == 1){
58     int ans = 1e9;
59     for(int i=2;i<=m;i++){ans = min(ans,abs(a[1][i]-a[1][i-1]));}
60     printf("%d
",ans);
61     return 0;
62     }
63     init();
64     divide();
65     return 0;
66 }