hdu 4183（最大流）

Pahom on Water

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

 

题意：有多个点，每个点给出坐标与半径，加入两个点相交，就可以从这两个点走。题目要求先从起点到终点，再从终点回到起点。从起点到终点的过程中，只能从频率小的走到频率大的点（前提是两点相交），从终点到起点的过程中，只能从频率大的走到频率小的。在走的过程中，除了起点与终点，别的只要走过就会消失，就是说只能走一次。问可不可以从起点到终点又回到起点。

解题思路：这个题我一开始的想法是拆点，但似乎并不能起到限制每个点只能走一次的作用，所以只好想别的办法了。之后我的想法就直接限制边，每条边的容量为1，这样跑最大流只要大于等于2，就一定是满足要求的。因为容量为1，代表着这一条边的两个顶点都只能经过一次，而最大流大于等于2是因为要从终点返回，那我们就可以认为从源点出发的几条流汇聚到汇点。此外，如果源点和汇点都能直接到达，就直接输出了。这道题只要想清了就很简单了。

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;const int maxn = 305;const int inf = 0x3f3f3f3f;struct Node{double f;int x,y,r;}p[maxn];struct Edge{int v,flow,next;}edge[maxn*maxn];int n,cnt,Head[maxn],layer[maxn];void addedge(int u,int v,int flow){edge[cnt].v = v;      edge[cnt].flow = flow;      edge[cnt].next = Head[u];      Head[u] = cnt++;        swap(u, v);        edge[cnt].v = v;      edge[cnt].flow = 0;      edge[cnt].next = Head[u];      Head[u] = cnt++;  }bool cmp(Node a,Node b){return a.f < b.f;}int dist(Node a,Node b){return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);}bool bfs(int start, int End)  {      queue<int> Q;Q.push(start);      memset(layer, 0, sizeof(layer));      layer[start] = 1;      while(Q.size())      {          int u = Q.front();Q.pop();          if(u == End) return true;          for(int j = Head[u]; j != -1; j = edge[j].next)          {              int v = edge[j].v;              if(layer[v] == 0 && edge[j].flow)              {                  layer[v] = layer[u] + 1;                  Q.push(v);              }          }      }      return false;  }  int dfs(int u, int MaxFlow, int End)  {      if(u == End) return MaxFlow;      int uflow = 0;      for(int j = Head[u]; j != -1; j = edge[j].next)      {          int v = edge[j].v;          if(layer[v] == layer[u] + 1 && edge[j].flow)          {              int flow = min(MaxFlow - uflow, edge[j].flow);              flow = dfs(v, flow, End);              edge[j].flow -= flow;              edge[j^1].flow += flow;              uflow += flow;              if(uflow == MaxFlow)break;          }      }      if(uflow == 0)          layer[u] = 0;      return uflow;  }  int dinic(int start, int End)  {      int MaxFlow = 0;      while(bfs(start, End) == true)          MaxFlow += dfs(start,inf,End);      return MaxFlow;  }  int main(){int t;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i = 1; i <= n; i++)scanf("%lf%d%d%d",&p[i].f,&p[i].x,&p[i].y,&p[i].r);memset(Head,-1,sizeof(Head));cnt = 0;sort(p+1,p+1+n,cmp);for(int i = 1; i < n; i++)for(int j = i + 1; j <= n; j++){if(dist(p[i],p[j]) < (p[i].r + p[j].r) * (p[i].r + p[j].r))addedge(i,j,1);}if(dist(p[1],p[n]) < (p[1].r + p[n].r) * (p[1].r + p[n].r))printf("Game is VALID\n");else{int ans = dinic(1,n);if(ans >= 2)printf("Game is VALID\n");else printf("Game is NOT VALID\n");}}return 0;}