题意:
给定一些冰块,每个冰块上有一些企鹅,每个冰块有一个可以跳出的次数限制,每个冰块位于一个坐标,现在每个企鹅跳跃力为d,问所有企鹅能否跳到一点上,如果可以输出所有落脚冰块,如果没有方案就打印-1
分析:
很显然的最大流问题。把每个冰块x拆成x和x',连x->x'流量为跳出的次数限制。枚举落脚冰块建图跑最大流即可
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
using namespace std;
const int maxn = 202 + 10;
const int INF = 1000000000;
struct Edge
{
int from,to,cap,flow;
};
struct Dinic
{
int n,m,s,t;
vector<Edge>edges;
vector<int>G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
void clearall(int n)
{
for(int i=0;i<n;i++)
G[i].clear();
edges.clear();
}
void clearflow()
{
for(int i=0;i<edges.size();i++)
edges[i].flow=0;
}
void addedge(int from,int to,int cap)
{
edges.push_back((Edge){from,to,cap,0});
edges.push_back((Edge){to,from,0,0});
m=edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool bfs()
{
memset(vis,0,sizeof(vis));
queue<int>Q;
Q.push(s);
vis[s]=1;
d[s]=0;
while(!Q.empty())
{
int x=Q.front();
Q.pop();
for(int i=0;i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(!vis[e.to]&&e.cap>e.flow)
{
vis[e.to]=1;
d[e.to]=d[x]+1;
Q.push(e.to);
}
}
}
return vis[t];
}
int dfs(int x,int a)
{
if(x==t||a==0)
{
return a;
}
int flow=0,f;
for(int& i=cur[x];i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(d[x]+1==d[e.to]&&(f=dfs(e.to,min(a,e.cap-e.flow)))>0)
{
e.flow+=f;
edges[G[x][i]^1].flow -= f;
flow+=f;
a-=f;
if(a==0)
break;
}
}
return flow;
}
int maxflow(int s,int t)
{
this->s=s;
this->t=t;
int flow=0;
//cout<<2<<endl;
while(bfs())
{
memset(cur,0,sizeof(cur));
flow+=dfs(s,INF);
}
//cout<<3<<endl;
return flow;
}
};
struct Node
{
int x,y,num;
}p[maxn];
Dinic solver;
double dis(Node a,Node b)
{
return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
vector<int>ans;
int main()
{
int t,n,total;
double D;
scanf("%d",&t);
while(t--)
{
scanf("%d%lf",&n,&D);
solver.clearall(2*n+1);
int s1=0;
total=0;
D=D*D;
for(int i=1;i<=n;i++)
{
int x,y,ni,cap;
scanf("%d%d%d%d",&x,&y,&ni,&cap);
p[i].x=x;p[i].y=y;p[i].num=ni;
total+=ni;
solver.addedge(s1,i,ni);
solver.addedge(i,i+n,cap);
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(i != j && D-dis(p[i],p[j])> 1e-6)
{
solver.addedge(i+n,j,INF);
}
int sum=0;
ans.clear();
//cout<<1<<endl;
for(int i=1;i<=n;i++)
{
solver.clearflow();
if(solver.maxflow(s1,i)==total)
{
ans.push_back(i);
sum++;
}
}
//cout<<sum<<endl;
if(sum==0)
printf("-1\n");
else
{
for(int i=0;i<ans.size()-1;i++)
printf("%d ",ans[i]-1);
printf("%d\n",ans[ans.size()-1]-1);
}
}
}
输入:
2
5 3.5
1 1 1 1
2 3 0 1
3 5 1 1
5 1 1 1
5 4 0 1
3 1.1
-1 0 5 10
0 0 3 9
2 0 1 1