Pressure DP Lightoj 1194 colored T-shirts

Give you a sequence of the minimum number of exchanges to put together the same numbers in this sequence

Range of numbers in a sequence 1-16

Number of numbers in a sequence 1-1e5

Ideas:

First preprocessing Cnt[i][j] indicates the number of interchanges to be used before the digit J is placed in the number I

Then enumerate each of the cases 1<<m

For example: M 0

000000000 the first I position is 0, indicating that the number I is disorderly, and vice versa is put good

The code is as follows

/********************************************author:crystalcreated time:file Name:* ********************* /#include <cstdio> #include <cstdlib> #include <iostream> #include < algorithm> #include <cstring> #include <climits> #include <string> #include <vector># Include <cmath> #include <stack> #include <queue> #include <set> #include <map>using Namespace Std;int cnt[20][20];int Pre[20];long long dp[1<<16];int nmin; #define INF 0x3f3f3f3fint main () {Freopen (" In.txt "," R ", stdin),//freopen (" OUT.txt "," w ", stdout), int t;cin >> t;for (int i=1;i<=t;i++) {cout <<" case  "<<" << I << ': ' << '; memset (cnt,0,sizeof cnt); memset (pre,0,sizeof pre); memset (dp,inf,sizeof DP) int n,m;cin >> n >> m;for (int j=0;j<n;j++) {int A; CIN >> a;a--;for (int r=0;r<m;r++) {Cnt[a][r] +=pre[r];//How many times a is required to change a before the R color//Add a number each time to update the front}pre[a]++;} for (int j=0;j<m;j++) for (int r=0;r<m;r++) cout << Cnt[j][r] << endl;dp[0]=0;int s = 1<<m;for (int j=0;j<s;j++)// 0 indicates a disorderly order for (int r=0;r<m;r++) {//r chaos, put all the disorderly order to the side if (J>>r & 1) continue;//if not 1continuelong long cost=0;for ( int k=0;k<m;k++) {if (K==r | | j>>k & 1) Continue;cost + = Cnt[r][k];} Dp[j | 1<<r]=min (dp[j | 1 << r],dp[j]+cost);} cout << dp[s-1] << ' \ n ';} return 0;}

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Pressure DP Lightoj 1194 colored T-shirts