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
-1 0
1 0
0 0
Sample Output
1 2
正解：半平面交。这就是个板子。
1 //It is made by wfj_2048~ 2 #include <algorithm> 3 #include <iostream> 4 #include <cstring> 5 #include <cstdlib> 6 #include <cstdio> 7 #include <vector> 8 #include <cmath> 9 #include <queue> 10 #include <stack> 11 #include <map> 12 #include <set> 13 #define inf (1LL<<40) 14 #define il inline 15 #define RG register 16 #define ll long long 17 18 using namespace std; 19 20 struct edge{ ll nt,to,dis; }g[1000010]; 21 22 ll no[110][110],f[110],head[30],cnt[30],dis[30],q[1000010],n,m,k,e,d,num; 23 24 il ll gi(){ 25 RG ll x=0,q=0; RG char ch=getchar(); while ((ch<'0' || ch>'9') && ch!='-') ch=getchar(); 26 if (ch=='-') q=1,ch=getchar(); while (ch>='0' && ch<='9') x=x*10+ch-48,ch=getchar(); return q ? -x : x; 27 } 28 29 il void insert(RG ll from,RG ll to,RG ll dis){ g[++num]=(edge){head[from],to,dis},head[from]=num; return; } 30 31 il ll spfa(){ 32 RG ll h=0,t=1; for (RG int i=2;i<=m;++i) dis[i]=inf; dis[1]=0,q[t]=1; 33 while (h<t){ 34 RG ll x=q[++h]; 35 for (RG ll i=head[x];i;i=g[i].nt){ 36 RG ll v=g[i].to; if (cnt[v]) continue; 37 if (dis[v]>dis[x]+g[i].dis) q[++t]=v,dis[v]=dis[x]+g[i].dis; 38 } 39 } 40 return dis[m]; 41 } 42 43 il void work(){ 44 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); } 45 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; } 46 for (RG ll i=1;i<=n;++i){ 47 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; } 48 RG ll l=spfa(); if (l<inf) f[i]=i*l; else f[i]=inf; 49 for (RG ll j=1;j<i;++j){ 50 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; } 51 RG ll l=spfa(); if (l<inf) f[i]=min(f[i],f[j]+(i-j)*l+k); 52 } 53 } 54 printf("%lld ",f[n]); return; 55 } 56 57 int main(){ 58 work(); 59 return 0; 60 }