题目链接:
BNUOJ 12887 - isumi (中文题面)
分析:
对题目中的图乱画一下发现,如果根据圆和矩形上下边相交的关系以及圆与圆之间相交的关系建图,由于存在一条从矩形左边走到右边且不与任何圆相交的路径等价于S和T不连通,那么只需选取尽可能少的点,使得删去这些点之后S和T不连通,拆点求最小割即可。
代码:
#include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<iostream> #include<algorithm> using namespace std; typedef long long ll; const int MAXN=10005; const int MAXM=100005; const int INF=0x3f3f3f3f; struct Edge { int to,next,cap,flow; }edge[MAXM]; int tol,head[MAXN]; int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN]; void init() { tol=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int w,int rw=0) { edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u]; edge[tol].flow=0; head[u]=tol++; edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v]; edge[tol].flow=0; head[v]=tol++; } int sap(int st,int ed,int N) { memset(gap,0,sizeof(gap)); memset(dep,0,sizeof(dep)); memcpy(cur,head,sizeof(head)); int u=st; pre[u]=-1; gap[0]=N; int ans=0; while(dep[st]<N) { if(u==ed) { int Min=INF; for(int i=pre[u];i!=-1;i=pre[edge[i^1].to]) if(Min>edge[i].cap-edge[i].flow) Min=edge[i].cap-edge[i].flow; for(int i=pre[u];i!=-1;i=pre[edge[i^1].to]) { edge[i].flow+=Min; edge[i^1].flow-=Min; } u=st; ans+=Min; continue; } bool flag=0; int v; for(int i=cur[u];i!=-1;i=edge[i].next) { v=edge[i].to; if(edge[i].cap-edge[i].flow && dep[v]+1==dep[u]) { flag=1; cur[u]=pre[v]=i; break; } } if(flag) { u=v; continue; } int Min=N; for(int i=head[u];i!=-1;i=edge[i].next) if(edge[i].cap-edge[i].flow && dep[edge[i].to]<Min) { Min=dep[edge[i].to]; cur[u]=i; } gap[dep[u]]--; if(!gap[dep[u]])return ans; dep[u]=Min+1; gap[dep[u]]++; if(u!=st)u=edge[pre[u]^1].to; } return ans; } struct Point { ll x,y; Point():x(0),y(0){} Point(ll x,ll y):x(x),y(y){} Point operator - (const Point &t)const { return Point(x-t.x,y-t.y); } ll len2() { return x*x+y*y; } }; struct Circle { Point o; ll r; Circle(ll x=0,ll y=0,ll r=0) { o=Point(x,y); this->r=r; } }c[105]; bool crossCircle(Circle c,ll y) { if(c.o.y-c.r<=y && c.o.y+c.r>=y)return 1; return 0; } bool interCircle(Circle a,Circle b) { ll d=(a.o-b.o).len2(); if(d>(a.r+b.r)*(a.r+b.r) || d<(a.r-b.r)*(a.r-b.r))return 0; return 1; } int main() { int l,w,n; scanf("%d%d%d",&l,&w,&n); ll x,y,d; for(int i=1;i<=n;i++) { scanf("%lld%lld%lld",&x,&y,&d); c[i]=Circle(x,y,d); } init(); for(int i=1;i<=n;i++)addedge(i,i+n,1); for(int i=1;i<=n;i++) { if(crossCircle(c[i],0))addedge(0,i,INF); if(crossCircle(c[i],w))addedge(i+n,2*n+1,INF); } for(int i=1;i<=n;i++) for(int j=i+1;j<=n;j++) if(interCircle(c[i],c[j])) { addedge(i+n,j,INF); addedge(j+n,i,INF); } printf("%d\n",sap(0,2*n+1,2*n+2)); return 0; }