HDU 3622 Bomb Game【2-SAT 问题】

HDU 3622 Bomb Game【2-SAT 问题】

分类： 【图论专辑】 2010-09-15 19:00  594人阅读  评论(0)  收藏  举报

       题目大意：有N对点，给定这N对点的坐标，现在要求找出一个最大的半径R，使得可以从每对点中选择一个点，并且这N个点以自己为圆心，半径为R的圆两两不相交。

       2-SAT问题解法：对于每一对点，我们且称之为“兄弟点”；对于这2*N个点，如果两个点之间“矛盾”，对于本题即两个圆相交，则连一条边。对这个图求强连通分量，然后看是否有某对点属于同一个强连通分量。如果N对点每一对都属于不同的强连通分量，满足。否则不满足。然后二分半径即可。

 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #define MAXN 210
 #define EPS 1e-3
 using namespace std;
 int n;
 struct Edge{
     int v,next;
     Edge(){}
     Edge(int _v,int _next):v(_v),next(_next){}
 }edge[MAXN*MAXN];
 int head[MAXN], size;
 inline void initial(){
     memset(head,-1,sizeof(int)*(n+2));
     size = 0;
 }
 inline void add_edge(int u,int v){
     edge[size] = Edge(v,head[u]);
     head[u] = size ++;
 }

 int belong[MAXN];
 bool vis[MAXN];
 int depth;
 int bcnt;
 int dfn[MAXN], low[MAXN];

 int stack[MAXN], tail ;
 inline void Tarjan(int u){
     dfn[u] = low[u] = depth ++;
     vis[u] = 1;
     stack[++tail] = u;
     for(int i = head[u];i != -1;i = edge[i].next){
         int v = edge[i].v;
         if(dfn[v] == -1){
             Tarjan(v);
             if(low[u] > low[v])
                 low[u] = low[v];
         }else if(vis[v] && low[u] > dfn[v]){
                 low[u] = dfn[v];
         }
     }
     if(low[u] == dfn[u]){
         bcnt ++;
         int j;
         do{
             j = stack[tail--];
             vis[j] = 0;
             belong[j] = bcnt;
         }while(j != u);
     }
 }
 double x[MAXN], y[MAXN];
 inline double Cal(int i,int j){
     return (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]);
 }
 inline bool can(){
     int N = n/2;
     for(int i = 1;i <= N; ++i)
         if(belong[i] == belong[i+N]) return false;
     return true;
 }
 int main()
 {
     int N;
     while(scanf("%d",&N) != EOF){
         n = 2*N;
         for(int i = 1;i <= N; ++i)
             scanf("%lf%lf%lf%lf",&x[i],&y[i],&x[i+N],&y[i+N]);
         double right = 10000.0*10000.0, left = 0.0, mid;
         while(right - left >= EPS){
             mid = (right + left)/2.0;
             initial();
             for(int i = 1;i < n; ++i)
                 for(int j = i + 1;j <= n; ++j){
                     //if(i + N == j) continue;
                     if(Cal(i,j) < mid){
                         int u, v;
                         if(i <= N) u = i + N;
                         else       u = i - N;
                         if(j <= N) v = j + N;
                         else       v = j - N;
                         add_edge(i,v);
                         add_edge(j,u);
                     }
                 }
             tail = depth = bcnt = 0;
             memset(vis,0,sizeof(bool)*(n+2));
             memset(dfn,-1,sizeof(int)*(n+2));
             memset(low,0,sizeof(int)*(n+2));
             for(int i = 1;i <= n; ++i)
                 if(dfn[i] == -1) Tarjan(i);
             if(can()) left = mid;
             else right = mid;
         }
         printf("%.2lf/n",sqrt(left)/2.0);
     }
 }