bzoj1007 [HNOI2008]水平可见直线

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Description

　　在xoy直角坐标平面上有n条直线L1,L2,...Ln,若在y值为正无穷大处往下看,能见到Li的某个子线段,则称Li为
可见的,否则Li为被覆盖的.
例如,对于直线:
L1:y=x; L2:y=-x; L3:y=0
则L1和L2是可见的,L3是被覆盖的.
给出n条直线,表示成y=Ax+B的形式(|A|,|B|<=500000),且n条直线两两不重合.求出所有可见的直线.

Input

　　第一行为N(0 < N < 50000),接下来的N行输入Ai,Bi

Output

　　从小到大输出可见直线的编号，两两中间用空格隔开,最后一个数字后面也必须有个空格

Sample Input

3
-1 0
1 0
0 0

Sample Output

1 2

正解：半平面交。这就是个板子。

//It is made by wfj_2048~#include <algorithm>#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <vector>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>#define inf (1LL<<40)#define il inline#define RG register#define ll long long using namespace std; struct edge{ ll nt,to,dis; }g[1000010]; ll no[110][110],f[110],head[30],cnt[30],dis[30],q[1000010],n,m,k,e,d,num; il ll gi(){    RG ll x=0,q=0; RG char ch=getchar(); while ((ch<'0' || ch>'9') && ch!='-') ch=getchar();    if (ch=='-') q=1,ch=getchar(); while (ch>='0' && ch<='9') x=x*10+ch-48,ch=getchar(); return q ? -x : x;} il void insert(RG ll from,RG ll to,RG ll dis){ g[++num]=(edge){head[from],to,dis},head[from]=num; return; } il ll spfa(){    RG ll h=0,t=1; for (RG int i=2;i<=m;++i) dis[i]=inf; dis[1]=0,q[t]=1;    while (h<t){    RG ll x=q[++h];    for (RG ll i=head[x];i;i=g[i].nt){        RG ll v=g[i].to; if (cnt[v]) continue;        if (dis[v]>dis[x]+g[i].dis) q[++t]=v,dis[v]=dis[x]+g[i].dis;    }    }    return dis[m];} il void work(){    n=gi(),m=gi(),k=gi(),e=gi(); for (RG ll i=1;i<=e;++i){ RG ll u=gi(),v=gi(),w=gi(); insert(u,v,w),insert(v,u,w); }    d=gi(); for (RG ll i=1;i<=d;++i){ RG ll id=gi(),a=gi(),b=gi(); for (RG ll j=a;j<=b;++j) no[id][j]=1; }    for (RG ll i=1;i<=n;++i){    memset(cnt,0,sizeof(cnt)); for (RG ll p=1;p<=m;++p) for (RG ll q=1;q<=i;++q) if (no[p][q]){ cnt[p]=1; break; }    RG ll l=spfa(); if (l<inf) f[i]=i*l; else f[i]=inf;    for (RG ll j=1;j<i;++j){        memset(cnt,0,sizeof(cnt)); for (RG ll p=1;p<=m;++p) for (RG ll q=j+1;q<=i;++q) if (no[p][q]){ cnt[p]=1; break; }        RG ll l=spfa(); if (l<inf) f[i]=min(f[i],f[j]+(i-j)*l+k);    }    }    printf("%lld\n",f[n]); return;} int main(){    work();    return 0;}

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