HDU 4183 Pahom on Water(最大流SAP)

Pahom on Water

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 629    Accepted Submission(s): 288

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

22 400.0 0 0 4789.0 7 0 24400.0 0 0 4789.0 7 0 2500.35 5 0 2500.32 5 0 3

 

Sample Output

Game is NOT VALIDGame is VALID

 

Source

The 2011 South Pacific Programming Contest

 

题意：就是给出n个pad（衬垫）接下来每一行代表一个pad，每个pad有一个颜色频率 f ，圆心坐标(x,y)和半径r. 现在从颜色频率f==400的 pad 开始走到 f==789.0的pad，这样走的规则是两圆有交点且f[u]<f[v]说明可以从u-->v。再从f==789.0走到f==400，走的规则是两圆有交点且f[u]>f[v]表明可以从u-->v。问有没有这样的一条路从起点走出后回到起点。（不能经过同一条边）

解题：因为从起点s走到终点t再回到s,不能经过同一条边。其实就是找两条从s-->t的路，经过的边不能相同，就可以满足要求。建图：衬垫 u,衬垫  v  ，如果 f[u]<f[v]且两圆有交点，则建一条边，边容为1。

#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>#include<math.h>using namespace std;#define captype intconst int MAXN = 100010;   //点的总数const int MAXM = 400010;    //边的总数const int INF = 1<<30;struct EDG{    int to,next;    captype cap,flow;} edg[MAXM];int eid,head[MAXN];int gap[MAXN];  //每种距离(或可认为是高度)点的个数int dis[MAXN];  //每个点到终点eNode 的最短距离int cur[MAXN];  //cur[u] 表示从u点出发可流经 cur[u] 号边int pre[MAXN];void init(){    eid=0;    memset(head,-1,sizeof(head));}//有向边 三个参数，无向边4个参数void addEdg(int u,int v,captype c,captype rc=0){    edg[eid].to=v; edg[eid].next=head[u];    edg[eid].cap=c; edg[eid].flow=0; head[u]=eid++;    edg[eid].to=u; edg[eid].next=head[v];    edg[eid].cap=rc; edg[eid].flow=0; head[v]=eid++;}captype maxFlow_sap(int sNode,int eNode, int n){//n是包括源点和汇点的总点个数，这个一定要注意    memset(gap,0,sizeof(gap));    memset(dis,0,sizeof(dis));    memcpy(cur,head,sizeof(head));    pre[sNode] = -1;    gap[0]=n;    captype ans=0;  //最大流    int u=sNode;    while(dis[sNode]<n){   //判断从sNode点有没有流向下一个相邻的点        if(u==eNode){   //找到一条可增流的路            captype Min=INF ;            int inser;            for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to])    //从这条可增流的路找到最多可增的流量Min            if(Min>edg[i].cap-edg[i].flow){                Min=edg[i].cap-edg[i].flow;                inser=i;            }            for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to]){                edg[i].flow+=Min;                edg[i^1].flow-=Min;  //可回流的边的流量            }            ans+=Min;            u=edg[inser^1].to;            continue;        }        bool flag = false;  //判断能否从u点出发可往相邻点流        int v;        for(int i=cur[u]; i!=-1; i=edg[i].next){            v=edg[i].to;            if(edg[i].cap-edg[i].flow>0 && dis[u]==dis[v]+1){                flag=true;                cur[u]=pre[v]=i;                break;            }        }        if(flag){            u=v;            continue;        }        //如果上面没有找到一个可流的相邻点，则改变出发点u的距离（也可认为是高度）为相邻可流点的最小距离+1        int Mind= n;        for(int i=head[u]; i!=-1; i=edg[i].next)        if(edg[i].cap-edg[i].flow>0 && Mind>dis[edg[i].to]){            Mind=dis[edg[i].to];            cur[u]=i;        }        gap[dis[u]]--;        if(gap[dis[u]]==0) return ans;  //当dis[u]这种距离的点没有了，也就不可能从源点出发找到一条增广流路径                                        //因为汇点到当前点的距离只有一种，那么从源点到汇点必然经过当前点，然而当前点又没能找到可流向的点，那么必然断流        dis[u]=Mind+1;//如果找到一个可流的相邻点，则距离为相邻点距离+1，如果找不到，则为n+1        gap[dis[u]]++;        if(u!=sNode) u=edg[pre[u]^1].to;  //退一条边    }    return ans;}struct node{    double f,x,y,r;}a[305];double ABS(double rr){    return rr>0?rr:-rr;}bool judge(int i,int j){    if(a[i].f>=a[j].f)        return 0;    double d=sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x) + (a[i].y-a[j].y)*(a[i].y-a[j].y));    if(a[i].r+a[j].r<d||d<ABS(a[i].r-a[j].r))        return 0;    return 1;}int main(){    int T,n;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        init();        int s , t;        for(int i=1; i<=n; i++){            scanf("%lf%lf%lf%lf",&a[i].f,&a[i].x,&a[i].y,&a[i].r);            if(a[i].f==400.0)                s=i;            else if(a[i].f==789.0)                t=i;        }        for(int i=1; i<=n; i++)            for(int j=1; j<=n; j++)            if(i!=j&&judge(i,j))                addEdg(i,j,1);         int ans = maxFlow_sap(s,t,n);         if(ans>1)            printf("Game is VALID\n");         else            printf("Game is NOT VALID\n");    }}