ACM Dance Recital 题解

 

题要求在所有表演的排列情况中， 找出换装次数最小的一种。 

考虑到题目数据规模比较小（n<=10, 10! = 3628800), 故可以暴力枚举所有的排列情况， 但有必要先将每两场表演所需换装的次数用二维数组记录下来， 否则可能会超时！

#include<bits/stdc++.h>using namespace std;int book[11][11];int book2[11];int book3[11];int minn;void dfs(int t, int n){if(t > n){int ans = 0;for(int i=1; i<=n-1; i++)ans += book[book2[i]][book2[i+1]];if(ans < minn) minn = ans;return;}for(int i=1; i<=n; i++){if(book3[i]==0){book2[t] = i;book3[i] = 1;dfs(t+1, n);book3[i] = 0;}}}int main(){int i,j,e,f,n,tmp;string s[30];cin>>n;for(i=1; i<=n; i++)cin>>s[i];for(i=1; i<=n; i++)for(j=1; j<=n; j++){tmp = 0;for(e=0; s[i][e]; e++)for(f=0; s[j][f]; f++)if(s[i][e] == s[j][f])tmp++;book[i][j] = tmp;}minn = INT_MAX;dfs(1, n);cout<<minn<<endl;return 0;}