【动态规划】【状态压缩DP】[UVa 1354]Mobile Computing

其实就是枚举一下每一次左边的用那些右边的用那些，处理一下当前所有组合的左端点和又短点就行了

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int MAXN = 6;
typedef pair<double, double> pii;
double sumw[(1<2];
int s;
double stone[MAXN+10];
double r;
int S;
vector vec[(1<2];
bool vis[(1<2];
int bitcount(int u){
    int ret = 0;
    while(u){
        if(u & 1) ret++;
        u >>= 1;
    }
    return ret;
}
void dfs(int ns){
    if(vis[ns]) return ;
    vis[ns] = true; vec[ns].clear();
    if(bitcount(ns) == 1){
        vec[ns].push_back(make_pair(0, 0));
        return ;
    }
    for(int s0=(ns-1)&ns; s0; s0=(s0-1)&ns){
        int s1 = ns ^ s0;
        dfs(s1); dfs(s0);
        int L1 = vec[s0].size(), L2 = vec[s1].size();
        for(int i=0;ifor(int j=0;jint main(){
    int T;
    scanf("%d", &T);
    while(T--){
        memset(sumw, 0, sizeof sumw);
        memset(vis, 0, sizeof vis);
        scanf("%lf%d", &r, &s);
        for(int i=0;iscanf("%lf", &stone[i]);
        S = 1 << s;
        for(int i=1;ifor(int j=0;jif(i&(1 << j)) sumw[i] += stone[j];
        dfs(S-1);
        double ans = -1;
        int Len1 = vec[S-1].size();
        for(int i=0;iif(vec[S-1][i].second - vec[S-1][i].first < r)
                if(vec[S-1][i].second - vec[S-1][i].first > ans)
                    ans = vec[S-1][i].second - vec[S-1][i].first;
        if(ans < 0) printf("-1\n");
        else printf("%.10lf\n", ans);
    }

    return 0;
}