POJ 3498 March of the Penguins（网络流+枚举）

Posted 2020-10-15

题目链接：http://poj.org/problem?id=3498

题目：

Description

Somewhere near the south pole, a number of penguins are standing on a number of ice floes. Being social animals, the penguins would like to get together, all on the same floe. The penguins do not want to get wet, so they have use their limited jump distance to get together by jumping from piece to piece. However, temperatures have been high lately, and the floes are showing cracks, and they get damaged further by the force needed to jump to another floe. Fortunately the penguins are real experts on cracking ice floes, and know exactly how many times a penguin can jump off each floe before it disintegrates and disappears. Landing on an ice floe does not damage it. You have to help the penguins find all floes where they can meet.

A sample layout of ice floes with 3 penguins on them.

Input

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

• One line with the integer N (1 ≤ N ≤ 100) and a floating-point number D (0 ≤ D ≤ 100 000), denoting the number of ice pieces and the maximum distance a penguin can jump.

• N lines, each line containing x_i, y_i, n_i and m_i, denoting for each ice piece its X and Y coordinate, the number of penguins on it and the maximum number of times a penguin can jump off this piece before it disappears (−10 000 ≤ x_i, y_i≤ 10 000, 0 ≤ n_i ≤ 10, 1 ≤ m_i ≤ 200).

Output

Per testcase:

• One line containing a space-separated list of 0-based indices of the pieces on which all penguins can meet. If no such piece exists, output a line with the single number −1.

Sample Input

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

Sample Output

1 2 4
-1

题意：n只企鹅跳冰块，每只最多跳d米,并且每只企鹅从当前冰块跳到另一个冰块上，当前冰块的寿命-1。问所有企鹅能否跳到同一块冰块上，如果可能的话，列举出所有可能性。
题解：注意题目中冰块下标是从0开始的。构图：源点S（0点），1-n枚举每个终点，把每个冰块所在的点分割成i（入口）和i+n（出口）两个点。S与i点连接，容量为该冰块上
原有企鹅的数目；i与i+n连接，容量为冰块的寿命；如果i冰块能够跳到j冰块，i+n和j连接，容量为INF。图构造好直接跑Dinic就可以啦。

  1 //POJ 3498
  2 #include <queue>
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <iostream>
  6 #include <algorithm>
  7 using namespace std;
  8 
  9 const int N=233;
 10 const int M=2*N*N;
 11 const int INF=0x3f3f3f3f;
 12 int n,s,t,cnt;
 13 int Head[N],Depth[N],cur[N],Map[N][N];
 14 int Next[M],V[M],W[M];
 15 
 16 void init(){
 17     cnt=-1;
 18     memset(Head,-1,sizeof(Head));
 19     memset(Next,-1,sizeof(Next));
 20 }
 21 
 22 void add_edge(int u,int v,int w){
 23     cnt++;Next[cnt]=Head[u];V[cnt]=v;W[cnt]=w;Head[u]=cnt;
 24     cnt++;Next[cnt]=Head[v];V[cnt]=u;W[cnt]=0;Head[v]=cnt;
 25 }
 26 
 27 bool bfs(){
 28     queue <int> Q;
 29     while(!Q.empty()) Q.pop();
 30     memset(Depth,0,sizeof(Depth));
 31     Depth[s]=1;
 32     Q.push(s);
 33     while(!Q.empty()){
 34         int u=Q.front();Q.pop();
 35         for(int i=Head[u];i!=-1;i=Next[i]){
 36             if((W[i]>0)&&(Depth[V[i]]==0)){
 37                 Depth[V[i]]=Depth[u]+1;
 38                 Q.push(V[i]);
 39             }
 40         }
 41     }
 42     if(Depth[t]==0) return 0;
 43     return 1;
 44 }
 45 
 46 int dfs(int u,int dist){
 47     if(u==t) return dist;
 48     for(int& i=cur[u];i!=-1;i=Next[i]){
 49         if((Depth[V[i]]==Depth[u]+1)&&W[i]!=0){
 50             int di=dfs(V[i],min(dist,W[i]));
 51             if(di>0){
 52                 W[i]-=di;
 53                 W[i^1]+=di;
 54                 return di;
 55             }
 56         }
 57     }
 58     return 0;
 59 }
 60 
 61 int Dinic(){
 62     int ans=0;
 63     while(bfs()){
 64         for(int i=0;i<=2*n;i++) cur[i]=Head[i];
 65         while(int d=dfs(s,INF)) ans+=d;
 66     }
 67     return ans;
 68 }
 69 
 70 struct TnT{
 71     double x,y;
 72     int n,m;
 73 }node[N];
 74 
 75 double dis(TnT a,TnT b){
 76     return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
 77 }
 78 
 79 int check(){
 80     for(int i=1;i<=n;i++) add_edge(s,i,node[i].n);
 81     for(int i=1;i<=n;i++) add_edge(i,i+n,node[i].m);
 82     for(int i=1;i<=n;i++)
 83     for(int j=1;j<=n;j++)
 84     if(Map[i][j]) add_edge(i+n,j,INF);
 85 
 86     return Dinic();
 87 
 88 }
 89 
 90 int main(){
 91     int T;
 92     double d;
 93     scanf("%d",&T);
 94     while(T--){
 95         int sum=0;
 96         scanf("%d %lf",&n,&d);
 97         for(int i=1;i<=n;i++){
 98             scanf("%lf %lf %d %d",&node[i].x,&node[i].y,&node[i].n,&node[i].m);
 99             sum+=node[i].n;
100         }
101         for(int i=1;i<=n;i++){
102             for(int j=1;j<=n;j++){
103                 if(i==j) Map[i][j]=1;
104                 else if(dis(node[i],node[j])<=d*d) Map[i][j]=1;
105                 else Map[i][j]=0;
106             }
107         }
108         vector <int> ans;
109         for(int i=1;i<=n;i++){
110             init();s=0;t=i;
111             if(check()==sum) ans.push_back(i);
112         }
113         int len=ans.size();
114         if(len==0) printf("-1\n");
115         else{
116             for(int i=0;i<len;i++){
117                 if(i!=len-1) printf("%d ",ans[i]-1);
118                 else if(i==len-1) printf("%d\n",ans[i]-1);
119             }
120         }
121 
122     }
123 
124     return 0;
125 }