UVA - 10911 Forming Quiz Teams 状态压缩
2014-11-20 23:45
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题目大意:有2*N个人,要求两人组成一对,因为两个人要经常一起练习,所以要求两个人的距离要近一些,求组成N个队伍时,每支队伍的距离和达到最短是多少
解题思路:因为最多只有20个人,所以可以用位运算符来表示人是否已经被匹配,dp[1<<i]表示i个人组成的队伍的距离和,则dp[1<<i] = min(dp[1<<i],solve((1<<i) ^(1<<j)^(1<<k))+dis[j][k]),1<<j和1<<k分别表示选中了第j个和第k个人组成一组,dis[j][k]表示第j个人到第k个人的距离,solve()是个递归调用函数,具体看代码
解题思路:因为最多只有20个人,所以可以用位运算符来表示人是否已经被匹配,dp[1<<i]表示i个人组成的队伍的距离和,则dp[1<<i] = min(dp[1<<i],solve((1<<i) ^(1<<j)^(1<<k))+dis[j][k]),1<<j和1<<k分别表示选中了第j个和第k个人组成一组,dis[j][k]表示第j个人到第k个人的距离,solve()是个递归调用函数,具体看代码
#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; #define maxn 66000 #define INF 0x3f3f3f3f int x[25],y[25]; double dp[maxn],dis[25][25]; int n; double solve(int s) { if(dp[s] != -1) return dp[s]; dp[s] = INF; for(int i = 0 ; i < n; i++) if((1<<i) &s) for(int j = 0; j < n; j++) if(i!=j &&((1<<j)&s)) dp[s] = min(dp[s],solve(s^(1<<i)^(1<<j))+dis[i][j]); return dp[s]; } int main() { int mark = 1; while(scanf("%d", &n) && n) { n <<= 1; char str[100]; for(int i = 0; i < n; i++) scanf("%s%d%d",str,&x[i],&y[i]); int end = (1 << n) - 1; for(int i = 0; i <= end; i++) dp[i] = -1; dp[0] = 0; for(int i = 0; i < n; i++) for(int j = 0; j < i; j++) dis[i][j] = dis[j][i] = sqrt(1.0*((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]))); solve(end); printf("Case %d: %.2lf\n",mark++,dp[end]); } }
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