uva10911 - Forming Quiz Teams(简单动归)

记忆化搜索+状态压缩

用二进制的01表示是否已分配的状态。

状态：dp[x]表示在状态x下所能达到最小的距离。

状态转移：dp[x] = min(dp[x^(1<<i)^(1<<j)]+dis(i....j));

代码如下：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 8
int n, vis[1<<2*N], x[2*N], y[2*N];
double dp[1<<2*N], f[2*N][2*N];
char str[25];
double dfs(int x)
{
    if(vis[x]) return dp[x];
    double &ans = dp[x];
    if(!x) {vis[x] = 1;  return ans = 0.0; }
    ans = 2000.0;
    for(int i = 0; i < 2*n; i++) if(x&(1<<i))
    for(int j = i+1; j < 2*n; j++) if(x&(1<<j))
    {
        ans = min(ans, dfs(x^(1<<i)^(1<<j))+f[i][j]);
    }
    vis[x] = 1;
    return ans;
}
int main ()
{
    int k = 0;
    while(scanf("%d",&n), n)
    {
        for(int i = 0; i < 2*n; i++)
            scanf("%s %d %d", str, &x[i], &y[i]);
        for(int i = 0; i < 2*n; i++) for(int j = i+1; j < 2*n; j++)
        f[i][j] = sqrt(1.0*(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
        int all = ((1<<2*n)-1);
        memset(vis,0,sizeof(vis));
        printf("Case %d: %.2lf\n",++k, dfs(all));
    }
    return 0;
}