【贪心】CodeChef PROTEPOI Protecting The Poison

题面在这里

其实很显然，

把所有蛇都投影到垂直的两个方向

对两个方向分别进行扫描线就可以了

示例程序：

#include<cstdio>#include<algorithm>using namespace std;const int maxn=100005,INF=0x3f3f3f3f;int tst,n,k,m,L,R,hx[maxn],hy[maxn],tx[maxn],ty[maxn],ans=0;struct data{    int s,t;    bool operator<(const data&b)const{        return s<b.s;    }}a[maxn];void work(){    int lst=L;    for (int i=1;i<=n&&lst<=R;){        if (lst<a[i].s) {ans-=INF;return;}        int far=-1;        while (i<=n&&a[i].s<=lst) far=max(far,a[i++].t);        lst=far+1;ans++;    }    if (lst<=R) ans-=INF;}int main(){    scanf("%d",&tst);    while (tst--){        scanf("%d%d%d",&n,&k,&m);        L=(n-k)/2+1;R=L+k-1;        for (int i=1;i<=m;i++){            scanf("%d%d%d%d",&hx[i],&hy[i],&tx[i],&ty[i]);            if (hx[i]==tx[i]){                if (hy[i]>ty[i]) swap(hy[i],ty[i]);                a[i].s=hy[i];a[i].t=ty[i];            }else a[i].s=a[i].t=hy[i];        }        sort(a+1,a+1+m);        ans=0;work();        for (int i=1;i<=m;i++)         if (hy[i]==ty[i]){            if (hx[i]>tx[i]) swap(hx[i],tx[i]);            a[i].s=hx[i];a[i].t=tx[i];         }else a[i].s=a[i].t=hx[i];        sort(a+1,a+1+m);        work();        printf("%d\n",ans<0?-1:ans);    }    return 0;}