poj 2296 Map Labeler(二分+2-sat判定)
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Map Labeler
Description
Map generation is a difficult task in cartography. A vital part of such task is automatic labeling of the cities in a map; where for each city there is text label to be attached to its location, so that no two labels overlap. In this problem, we are concerned with a simple case of automatic map labeling.
Assume that each city is a point on the plane, and its label is a text bounded in a square with edges parallel to x and y axis. The label of each city should be located such that the city point appears exactly in the middle of the top or bottom edges of the label. In a good labeling, the square labels are all of the same size, and no two labels overlap, although they may share one edge. Figure 1 depicts an example of a good labeling (the texts of the labels are not shown.) Given the coordinates of all city points on the map as integer values, you are to find the maximum label size (an integer value) such that a good labeling exists for the map. 
Input
The first line contains a single integer t (1 <= t <= 10), the number of test cases. Each test case starts with a line containing an integer m (3 ≤ m ≤ 100), the number of cities followed by m lines of data each containing a pair of integers; the first integer (X) is the x and the second one (Y) is the y coordinates of one city on the map (-10000 ≤X, Y≤ 10000). Note that no two cities have the same (x, y) coordinates.
Output
The output will be one line per each test case containing the maximum possible label size (an integer value) for a good labeling.
Sample Input 1 6 1 1 2 3 3 2 4 4 10 4 2 5 Sample Output 2 Source
Tehran 2003
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题目：http://poj.org/problem?id=2296
题意：给你n个点，要你在这n个点上方一个正方形，点只能在正方形的上边或下边的中点上，所有正方形大小一样，不能重叠，求最大的正方形。。。
分析：跟HDU 3622 Bomb Game差不多，只是变成正方形判定而已，分清楚各种状态就行
PS:强连通老是写错那么几个变量，调试好久啊= =，不过还是1Y了
代码：
#include<cstdio>
#include<iostream>
using namespace std;
const int mm=88888;
const int mn=222;
int ver[mm],next[mm];
int x[mn],y[mn],head[mn],dfn[mn],low[mn],q[mn],id[mn];
int i,j,k,l,r,m,n,idx,top,cnt,edge,t,ans;
void add(int u,int v)
{
ver[edge]=v,next[edge]=head[u],head[u]=edge++;
}
void dfs(int u)
{
dfn[u]=low[u]=++idx;
q[top++]=u;
for(int i=head[u],v;i>=0;i=next[i])
if(!dfn[v=ver[i]])
dfs(v),low[u]=min(low[u],low[v]);
else if(!id[v])low[u]=min(low[u],dfn[v]);
if(low[u]==dfn[u])
{
id[u]=++cnt;
while(q[--top]!=u)id[q[top]]=cnt;
}
}
void Tarjan()
{
for(idx=top=cnt=i=0;i<n+n;++i)dfn[i]=id[i]=0;
for(i=0;i<n+n;++i)
if(!dfn[i])dfs(i);
}
int abc(int x)
{
return x<0?-x:x;
}
bool ok()
{
for(edge=i=0;i<n+n;++i)head[i]=-1;
for(i=0;i<n;++i)
for(j=i+1;j<n;++j)
if(abc(x[i]-x[j])<m&&abc(y[i]-y[j])<2*m)
{
if(abc(y[i]-y[j])>=m)
{
if(y[i]<y[j])
{
add(i<<1|1,j<<1|1);
add(j<<1,i<<1);
}
else
{
add(i<<1,j<<1);
add(j<<1|1,i<<1|1);
}
}
else if(y[i]<y[j])add(i<<1|1,i<<1),add(j<<1,j<<1|1);
else if(y[i]>y[j])add(i<<1,i<<1|1),add(j<<1|1,j<<1);
else
{
add(i<<1|1,j<<1),add(i<<1,j<<1|1);
add(j<<1,i<<1|1),add(j<<1|1,i<<1);
}
}
Tarjan();
for(i=0;i<n+n;i+=2)
if(id[i]==id[i^1])return 0;
return 1;
}
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=0;i<n;++i)
scanf("%d%d",&x[i],&y[i]);
ans=l=0,r=1e5;
while(l<=r)
{
m=(l+r)>>1;
if(ok())ans=m,l=m+1;
else r=m-1;
}
printf("%d\n",ans);
}
return 0;
}