題意：二維平面有一堆點，每個點有價值v和刪掉這個點能得到的長度l，問你刪掉最少的價值能把剩余點圍起來，價值一樣求刪掉的點最少

思路：n<=15，那麽直接遍歷2^15，判斷每種情況。這裏要優化一下，如果價值比當前最優大了continue。POJ的G++輸出要用%f...orz，還是乖乖用C++...

代碼：

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include
 
<string>
#include<cstdio>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
using namespace std;
const int maxn = 100 + 10;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
struct node{
    double x, y, l;
    
 
int v, id;
}p[maxn], s[maxn], q[maxn];
int n, top;
double dis(node a, node b){
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
bool cmp(node a, node b){
    double A = atan2((a.y - p[1].y), (a.x - p[1].x));
    double B = atan2((b.y - p[1].y), (b.x - p[1].x));
    
 
if(A != B) return A < B;
    else{
        return dis(a, p[1]) < dis(b, p[1]);
    }
}
double cross(node a, node b, node c){   //(a->b)X(a->c)
    return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
}
void solve(){
    int pos = 1;
    for(int i = 2; i <= n; i++){
        if(p[i].y < p[pos].y || (p[i].y == p[pos].y && p[i].x < p[pos].x)){
            pos = i;
        }
    }
    swap(p[1], p[pos]);
    sort(p + 2, p + n + 1, cmp);
    s[0] = p[1], s[1] = p[2];
    top = 1;
    for(int i = 3; i <= n; i++){
        while(top >= 1 && cross(s[top - 1], p[i], s[top]) >= 0){
            top--;
        }
        s[++top] = p[i];
    }
}
double need(double len){
    if(n <= 1) return len;
    if(n == 2){
        return len - 2.0 * dis(p[1], p[2]);
    }
    solve();
    double Need = 0;
    for(int i = 0; i < top; i++){
        Need += dis(s[i], s[i + 1]);
    }
    Need += dis(s[top], s[0]);
    return len - Need;
}
int main(){
    int N, ca = 1;
    while(~scanf("%d", &N) && N){
        for(int i = 0; i < N; i++){
            scanf("%lf%lf%d%lf", &q[i].x, &q[i].y, &q[i].v, &q[i].l);
            q[i].id = i + 1;
        }
        int sz = 0, que[20], tempQue[20];
        int MinValue = INF;
        double ans = 0;
        for(int i = 1; i < (1 << N); i++){
            int num = 0, value = 0;
            double len = 0;
            n = 0;
            for(int j = 0; j < N; j++){
                if(i & (1 << j)){
                    p[++n] = q[j];
                }
                else{
                    value += q[j].v;
                    len += q[j].l;
                    tempQue[num] = q[j].id;
                    num++;
                }
            }
            if(value > MinValue) continue;
            double tmp = need(len);
            if(tmp < 0) continue;
            if(value < MinValue || (value == MinValue && num < sz)){
                MinValue = value;
                ans = tmp;
                sz = num;
                for(int j = 0; j < num; j++){
                    que[j] = tempQue[j];
                }
            }
        }
        if(ca != 1) printf("\n");
        printf("Forest %d\n", ca++);
        printf("Cut these trees: ");
        for(int i = 0; i < sz; i++){
            printf("%d ", que[i]);
        }
        printf("\n");
        printf("Extra wood: %.2lf\n", ans);
    }
    return 0;
}

POJ 1873 The Fortified Forest（凸包）題解