HDU 3622 Bomb Game (二分+2-SAT)

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题目地址：HDU 3622

先二分半径，然后小于该半径的不能选，对这些不能选的点对进行加边。然后判断可行性即可。

代码如下：

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>#include <set>#include <algorithm>using namespace std;#define LL __int64const int INF=0x3f3f3f3f;const double eqs=1e-3;int head[410], cnt, top, ans, index;int dfn[410], low[410], instack[410], stak[410], belong[410];struct node{    int u, v, next;}edge[1000000];struct Point{    int x, y;}dian[10000];void add(int u, int v){    edge[cnt].v=v;    edge[cnt].next=head[u];    head[u]=cnt++;}void tarjan(int u){    dfn[u]=low[u]=++index;    instack[u]=1;    stak[++top]=u;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(!dfn[v])        {            tarjan(v);            low[u]=min(low[u],low[v]);        }        else if(instack[v])        {            low[u]=min(low[u],dfn[v]);        }    }    if(dfn[u]==low[u])    {        ans++;        while(1)        {            int v=stak[top--];            instack[v]=0;            belong[v]=ans;            if(u==v) break;        }    }}void init(){    memset(head,-1,sizeof(head));    cnt=0;    memset(dfn,0,sizeof(dfn));    memset(instack,0,sizeof(instack));    top=ans=index=0;}double dist(Point x, Point y){    return sqrt((x.x-y.x)*(x.x-y.x)*1.0+(x.y-y.y)*(x.y-y.y));}int solve(double mid, int n){    int i, j;    init();    for(i=0;i<n<<1;i++)    {        for(j=0;j<i;j++)        {            if(dist(dian[i],dian[j])<mid)            {                add(i,j^1);                add(j,i^1);                //add(j^1,i);                //add(i^1,j);                //printf("%d %d %d %d\n",i<<1,j<<1|1,j<<1,i<<1|1);            }        }    }    for(i=0;i<n<<1;i++)    {        if(!dfn[i])            tarjan(i);    }    for(i=0;i<n;i++)    {        if(belong[i<<1]==belong[i<<1|1])            return 0;    }    return 1;}int main(){    int n, i, j;    while(scanf("%d",&n)!=EOF)    {        for(i=0;i<n;i++)        {            scanf("%d%d%d%d",&dian[i<<1].x,&dian[i<<1].y,&dian[i<<1|1].x,&dian[i<<1|1].y);        }        double r=40000.0, l=0, mid, ans;        while(r-l>eqs)        {            mid=(l+r)/2;            //printf("%.2lf\n",mid);            if(solve(mid, n))            {                ans=mid;                l=mid;            }            else                r=mid;        }        printf("%.2lf\n",ans/2.0);    }    return 0;}

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