UVa 1354 枚举子集 Mobile Computing

只要枚举左右两个子天平砝码的集合，我们就能算出左右两个悬挂点到根悬挂点的距离。

但是题中要求找尽量宽的天平但是不能超过房间的宽度，想不到要怎样记录结果。

参考别人代码，用了一个结构体的vector，保存每个集合合法方案的左右两端最长的距离。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#include <cmath>
#define MP make_pair
#define Ft first
#define Sd second
using namespace std;

typedef pair<double, double> PDD;

const int maxn = 10;
const int maxs = 1000;

vector<PDD> tree[maxs];

int n;
double r;
double a[maxn], w[maxs];
bool vis[maxs];

int bitcount(int x)
{
    int ans = 0;
    while(x) { ans += (x & 1); x >>= 1; }
    return ans;
}

void dfs(int S)
{
    if(vis[S]) return ;
    vis[S] = true;
    if(bitcount(S) == 1) { tree[S].push_back(MP(0, 0)); return ; }

    PDD t = MP(0, 0);
    for(int s1 = (S-1)&S; s1; s1 = (s1-1)&S)
    {
        int s2 = S ^ s1;
        dfs(s1); dfs(s2);
        double x1 = w[s2] / w[S], x2 = w[s1] / w[S];
        for(int i = 0; i < tree[s1].size(); i++)
            for(int j = 0; j < tree[s2].size(); j++)
            {
                t.Ft = max(x1 + tree[s1][i].Ft, tree[s2][j].Ft - x2);
                t.Sd = max(x2 + tree[s2][j].Sd, tree[s1][i].Sd - x1);
                if(t.Ft + t.Sd < r) tree[S].push_back(t);
            }
    }
}

int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        scanf("%lf%d", &r, &n);
        for(int i = 0; i < n; i++) scanf("%lf", a + i);
        int all = (1 << n) - 1;

        for(int i = 0; i <= all; i++)
        {
            w[i] = 0;
            tree[i].clear();
            for(int j = 0; j < n; j++) if(i & (1 << j))
                w[i] += a[j];
        }

        memset(vis, false, sizeof(vis));
        dfs(all);
        double ans = -1;
        for(int i = 0; i < tree[all].size(); i++) ans = max(ans, tree[all][i].Ft + tree[all][i].Sd);
        if(ans < 0) puts("-1");
        else printf("%.9f
", ans);
    }

    return 0;
}