Pahom on Water(最大流)

Pahom on Water

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 770    Accepted Submission(s): 353

Problem Description

Pahom on Water is an interactive computer game inspired by a short story of Leo Tolstoy about a poor man who, in his lust for land, forfeits everything. The game's starting screen displays a number of circular pads painted with colours from the visible light spectrum. More than one pad may be painted with the same colour (defined by a certain frequency) except for the two colours red and violet. The display contains only one red pad (the lowest frequency of 400 THz) and one violet pad (the highest frequency of 789 THz). A pad may intersect, or even contain another pad with a different colour but never merely touch its boundary. The display also shows a figure representing Pahom standing on the red pad. The game's objective is to walk the figure of Pahom from the red pad to the violet pad and return back to the red pad. The walk must observe the following rules: 1.If pad α and pad β have a common intersection and the frequency of the colour of pad α is strictly smaller than the frequency of the colour of pad β, then Pahom figure can walk from α to β during the walk from the red pad to the violet pad 2. If pad α and pad β have a common intersection and the frequency of the colour of pad α is strictly greater than the frequency of the colour of pad β, then Pahom figure can walk from α to β during the walk from the violet pad to the red pad 3. A coloured pad, with the exception of the red pad, disappears from display when the Pahom figure walks away from it. The developer of the game has programmed all the whizzbang features of the game. All that is left is to ensure that Pahom has a chance to succeed in each instance of the game (that is, there is at least one valid walk from the red pad to the violet pad and then back again to the red pad.) Your task is to write a program to check whether at least one valid path exists in each instance of the game.

 

Input

The input starts with an integer K (1 <= K <= 50) indicating the number of scenarios on a line by itself. The description for each scenario starts with an integer N (2 <= N <= 300) indicating the number of pads, on a line by itself, followed by N lines that describe the colors, locations and sizes of the N pads. Each line contains the frequency, followed by the x- and y-coordinates of the pad's center and then the radius. The frequency is given as a real value with no more than three decimal places. The coordinates and radius are given, in meters, as integers. All values are separated by a single space. All integer values are in the range of -10,000 to 10,000 inclusive. In each scenario, all frequencies are in the range of 400.0 to 789.0 inclusive. Exactly one pad will have a frequency of “400.0” and exactly one pad will have a frequency of “789.0”.

 

Output

The output for each scenario consists of a single line that contains: Game is VALID, or Game is NOT VALID

 

Sample Input

2
2
400.0 0 0 4
789.0 7 0 2
4
400.0 0 0 4
789.0 7 0 2
500.35 5 0 2
500.32 5 0 3

 

Sample Output

Game is NOT VALID
Game is VALID

题解：题意就是光圈相交了才能走，只能从频率大的到小的，问是否能从红光到紫光再回到红光，红光到紫光频率要从小到大<<<<紫光到红光频率要从大到小>>>>>,由此直接红光与源点连，紫光连汇点，频率从小到大，就加边，能从红到紫，自然能从紫到红了；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
const int INF=0x3f3f3f3f;
const int MAXN=;
const int MAXM=<<;
struct Edge{
    int from,to,next,cup,flow;
};
Edge edg[MAXM];
int head[MAXM],dis[MAXN],edgnum,vis[MAXN],h[MAXM];
void initial(){
    edgnum=;mem(head,-);
}
struct Dot{
    double fq,x,y,r;
}dt[];
void add(int u,int v,int w){
    Edge E={u,v,head[u],w,};
    edg[edgnum]=E;
    head[u]=edgnum++;
    E={v,u,head[v],,};
    edg[edgnum]=E;
    head[v]=edgnum++;
}
bool bfs(int s,int e){
    mem(dis,-);mem(vis,);
    queue<int>dl;
    dis[s]=;vis[s]=;dl.push(s);
    while(!dl.empty()){
        int u=dl.front();
        dl.pop();
        for(int i=head[u];i!=-;i=edg[i].next){
            Edge v=edg[i];
            if(!vis[v.to]&&v.cup-v.flow){
                vis[v.to]=;
                dis[v.to]=dis[u]+;
                dl.push(v.to);//
                if(v.to==e)return true;
            }
        }
    }
    return false;
}
int dfs(int x,int la,int e){

    if(x==e||la==)return la;
    int temp,flow=;
    for(int i=h[x];i!=-;i=edg[i].next){
        Edge &v=edg[i];
        if(dis[v.to]==dis[x]+&&(temp=dfs(v.to,min(la,v.cup-v.flow),e))>){
            flow+=temp;
            v.flow+=temp;
            edg[i^].flow-=temp;
            la-=temp;
            if(la==)break;
        }
    }
    return flow;
}
int maxflow(int s,int e){
    int flow=;
    while(bfs(s,e)){
        memcpy(h,head,sizeof(head));
        flow+=dfs(s,INF,e);
    }
    return flow;
}
bool judge(Dot a,Dot b){
    if(a.fq<b.fq&&(pow(b.x-a.x,)+pow(b.y-a.y,)<pow(b.r+a.r,)))return true;
     return false;
}
int main(){
    int k,N;
    scanf("%d",&k);
    while(k--){
        initial();
        scanf("%d",&N);
        for(int i=;i<=N;i++){
            scanf("%lf%lf%lf%lf",&dt[i].fq,&dt[i].x,&dt[i].y,&dt[i].r);
            if(fabs(789.0-dt[i].fq)<=1e-)add(i,N+,);
            if(fabs(dt[i].fq-400.0)<=1e-)add(,i,);
        }
        for(int i=;i<=N;i++){
            for(int j=;j<=N;j++){
                if(i==j)continue;
                if(judge(dt[i],dt[j]))add(i,j,);
            }
        }
        if(maxflow(,N+)==)puts("Game is VALID");
        else puts("Game is NOT VALID");
    }
    return ;
}