poj 1873 The Fortified Forest(凸包+枚举)

【题目大意】:给出n棵树,每棵树有坐标,权值,和长度,现在要砍下其中的某些树,来做围栏来把剩下的树围起来,问损失的数最小价值是多少。


【解题思路】:明显求围栏就是一个凸包问题,由于树最多只有13棵,所以我们可以直接枚举哪些树留下,来求凸包,取最小值即可。。


【代码】:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <cctype>
#include <map>
#include <iomanip>
                   
using namespace std;
                   
#define eps 1e-8
#define PI acos(-1.0)
#define inf 1<<30
#define linf 1LL<<60
#define pb push_back
#define lc(x) (x << 1)
#define rc(x) (x << 1 | 1)
#define lowbit(x) (x & (-x))
#define ll long long
#define INF 999999999.9

struct Point{
    double x, y, dis ,val, len;
}tmp[20],pt[20],stack[20],p0;//stack为凸包上的点
int top,tot;
int n,mask;
double len,lens;

//计算几何距离
double get_dis(double x1, double y1, double x2, double y2){
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

//极角比较, 返回-1: p0p1在p0p2的右侧,返回0:p0,p1,p2共线
int Cmp_PolarAngel(struct Point p1, struct Point p2, struct Point pb){
    double delta=(p1.x-pb.x)*(p2.y-pb.y)-(p2.x-pb.x)*(p1.y-pb.y);
    if (delta<0.0) return 1;
    else if (delta==0.0) return 0;
    else return -1;
}

// 判断向量p2p3是否对p1p2构成左旋
bool Is_LeftTurn(struct Point p3, struct Point p2, struct Point p1){
    int type=Cmp_PolarAngel(p3, p1, p2);
    if (type<0) return true;
    return false;
}

//先按极角排,再按距离由小到大排
int Cmp(const void*p1, const void*p2){
    struct Point*a1=(struct Point*)p1;
    struct Point*a2=(struct Point*)p2;
    int type=Cmp_PolarAngel(*a1, *a2, p0);
    if (type<0) return -1;
    else if (type==0){
        if (a1->dis<a2->dis) return -1;
        else if (a1->dis==a2->dis) return 0;
        else return 1;
    }
    else return 1;
}

//求凸包
void Solve(int n,double len){
    int k;
    p0.x=p0.y=INF;
    for (int i=0; i<n; i++){
        if (pt[i].y < p0.y){
            p0.y=pt[i].y;
            p0.x=pt[i].x;
            k=i;
        }
        else if (pt[i].y==p0.y){
            if (pt[i].x<p0.x){
                p0.x=pt[i].x;
                k=i;
            }
        }
    }
    pt[k]=pt[0];
    pt[0]=p0;
    for (int i=1; i<n; i++)
        pt[i].dis=get_dis(pt[i].x,pt[i].y,p0.x,p0.y);
    qsort(pt+1, n-1, sizeof(struct Point), Cmp);
    //去掉极角相同的点
    tot=1;
    for (int i=2; i<n; i++)
        if (Cmp_PolarAngel(pt[i], pt[i-1], p0))
            pt[tot++]=pt[i-1];
    pt[tot++]=pt[n-1];
    //求凸包
    top=1;
    stack[0]=pt[0];
    stack[1]=pt[1];
    for (int i=2; i<tot; i++){
        while (top>=1 && Is_LeftTurn(pt[i],stack[top],stack[top-1])==false)
            top--;
        stack[++top]=pt[i];
    }
    double ans=0;
    double len1=0;
    stack[top+1]=stack[0];
    for (int i=1; i<=top+1; i++) len1+=get_dis(stack[i-1].x,stack[i-1].y,stack[i].x,stack[i].y);
    lens=len1;    
}

int main(){
    int Cas=0;
    bool flag=true;
    while (~scanf("%d",&n)){
        if (flag) flag=false;
        else cout << endl;
        if (n==0) break;
        for (int i=0; i<n; i++) scanf("%lf%lf%lf%lf",&tmp[i].x,&tmp[i].y,&tmp[i].val,&tmp[i].len);
        double ans=INF;
        double alen;
        int mask;
        for (int i=0; i<(1<<n); i++){
            memset(pt,0,sizeof(pt));
            int m=-1;
            double len=0.0;
            double tmp1=0;
            for (int j=0; j<n; j++)
                if (i&(1<<j)) m++,pt[m]=tmp[j];
                else tmp1+=tmp[j].val,len+=tmp[j].len;

            if (m+1>=1) {
                Solve(m+1,len); 
                if (ans>tmp1 && tmp1!=0 && lens<=len) {alen=len-lens; ans=tmp1; mask=i;}
            }
        }
        Cas++;
        cout << "Forest " << Cas <<endl;
        cout << "Cut these trees:";
        for (int i=0; i<n; i++) 
            if (!(mask&(1<<i))) printf(" %d",i+1);
        cout << endl;
        printf("Extra wood: %.2f\n",alen);
    }
    return 0;
}


你可能感兴趣的:(poj 1873 The Fortified Forest(凸包+枚举))