POJ 3498 March of the Penguins(枚举+最大流)

题意：在X,Y坐标系中有N(N<=100)个冰块...有些冰块上有若干只企鹅..每只企鹅一次最多跳M距离..一个冰块在有Mi个企鹅离开..就会消失..问有哪些冰块可以作为集合点..就是所有企鹅都能成功到这个冰块上来.

思路：枚举每一块冰块，看看最大流能否等于企鹅总数即可

           建图：把每块冰分成两个点i和i+n. i表示进入i冰块的点(可以有无数企鹅过来,所以从别的冰到i有边,容量为INF) i+n表示从i冰块出去的点(最多只能有Mi企鹅从这跳出去,所以从i到i+n有边,且容量为Mi)

                     从源点S到i有边(S, i, i点初始企鹅数).

                     从i到i+n有边(i, i+n, Mi). 表示第i块冰最多只有Mi个企鹅能跳走.

                     因为i+n表示的是第i个跳走的点,所以如果冰块i和j之间的距离<=企鹅能跳跃的距离M,有边(i+n, j, INF)

                     假设我们当前枚举第x块冰块作为集合点,那么(x分成x和x+n两个点)x点就是汇点(不是x+n点哦),我们只要计算到x点的流量是否==企鹅总数即可.

Trick：如果有拆点的话在初始化的时候记得修改初始化函数..这个傻逼错误我调了一个多小时...

#include <cstdio>#include <queue>#include <cstring>#include <iostream>#include <cstdlib>#include <algorithm>#include <vector>#include <map>#include <string>#include <set>#include <ctime>#include <cmath>#include <cctype>using namespace std;#define maxn 550#define INF 1e9#define LLINF 1LL<<60#define LL long longint cas=1,T;struct Edge{int from,to,cap,flow;Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}};int n,m;struct Dinic{//int n,m;    int s,t;vector<Edge>edges;        //边数的两倍vector<int> G[maxn];      //邻接表，G[i][j]表示结点i的第j条边在e数组中的序号bool vis[maxn];           //BFS使用int d[maxn];              //从起点到i的距离int cur[maxn];            //当前弧下标void init(){   for (int i=0;i<=2*n+2;i++)                   //初始化！！！如果有拆点记得这里要改   G[i].clear();   edges.clear();  // memset(d,0,sizeof(d));}void AddEdge(int from,int to,int cap){edges.push_back(Edge(from,to,cap,0));edges.push_back(Edge(to,from,0,0));        //反向弧int mm=edges.size();G[from].push_back(mm-2);G[to].push_back(mm-1);}bool BFS(){memset(vis,0,sizeof(vis));queue<int>q;q.push(s);d[s]=0;vis[s]=1;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;while (BFS()){memset(cur,0,sizeof(cur));flow+=DFS(s,INF);}return flow;}}dc;int sum;double limit;struct Node{double x,y;int now;      //有多少只企鹅在冰上int num;      //一块冰可以让多少只企鹅跳}pie[maxn];double getdist(Node a,Node b){return ((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}bool solve(int ice){dc.init();    for (int i = 1;i<=n;i++){if (pie[i].now)  dc.AddEdge(0,i,pie[i].now);}for (int i = 1;i<=n;i++){dc.AddEdge(i,i+n,pie[i].num);}for (int i = 1;i<=n;i++)for (int j = i+1;j<=n;j++)if (getdist(pie[i],pie[j]) <= limit*limit){   dc.AddEdge(i+n,j,INF);               dc.AddEdge(j+n,i,INF);}return dc.Maxflow(0,ice) == sum;}int main(){//freopen("in","r",stdin);scanf("%d",&T);while (T--){scanf("%d%lf",&n,&limit);sum=0;for (int i = 1;i<=n;i++){scanf("%lf%lf%d%d",&pie[i].x,&pie[i].y,&pie[i].now,&pie[i].num);sum+=pie[i].now;                 //统计企鹅的数量}vector <int> ans;ans.clear();        for (int i = 1;i<=n;i++){if (solve(i))ans.push_back(i);}if (ans.size()==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);}}//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);return 0;} 

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

25 3.51 1 1 12 3 0 13 5 1 15 1 1 15 4 0 13 1.1-1 0 5 100 0 3 92 0 1 1

Sample Output

1 2 4-1