poj 2749 Building roads （二分+拆点+2-sat）
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Building roads
Description
Farmer John's farm has N barns, and there are some cows that live in each barn. The cows like to drop around, so John wants to build some roads to connect these barns. If he builds roads for every pair of different barns, then he must build N * (N - 1) / 2 roads, which is so costly that cheapskate John will never do that, though that's the best choice for the cows.
Clever John just had another good idea. He first builds two transferring point S1 and S2, and then builds a road connecting S1 and S2 and N roads connecting each barn with S1 or S2, namely every barn will connect with S1 or S2, but not both. So that every pair of barns will be connected by the roads. To make the cows don't spend too much time while dropping around, John wants to minimize the maximum of distances between every pair of barns. That's not the whole story because there is another troublesome problem. The cows of some barns hate each other, and John can't connect their barns to the same transferring point. The cows of some barns are friends with each other, and John must connect their barns to the same transferring point. What a headache! Now John turns to you for help. Your task is to find a feasible optimal road-building scheme to make the maximum of distances between every pair of barns as short as possible, which means that you must decide which transferring point each barn should connect to. We have known the coordinates of S1, S2 and the N barns, the pairs of barns in which the cows hate each other, and the pairs of barns in which the cows are friends with each other. Note that John always builds roads vertically and horizontally, so the length of road between two places is their Manhattan distance. For example, saying two points with coordinates (x1, y1) and (x2, y2), the Manhattan distance between them is |x1 - x2| + |y1 - y2|. Input
The first line of input consists of 3 integers N, A and B (2 <= N <= 500, 0 <= A <= 1000, 0 <= B <= 1000), which are the number of barns, the number of pairs of barns in which the cows hate each other and the number of pairs of barns in which the cows are friends with each other.
Next line contains 4 integer sx1, sy1, sx2, sy2, which are the coordinates of two different transferring point S1 and S2 respectively. Each of the following N line contains two integer x and y. They are coordinates of the barns from the first barn to the last one. Each of the following A lines contains two different integers i and j(1 <= i < j <= N), which represent the i-th and j-th barns in which the cows hate each other. The same pair of barns never appears more than once. Each of the following B lines contains two different integers i and j(1 <= i < j <= N), which represent the i-th and j-th barns in which the cows are friends with each other. The same pair of barns never appears more than once. You should note that all the coordinates are in the range [-1000000, 1000000]. Output
You just need output a line containing a single integer, which represents the maximum of the distances between every pair of barns, if John selects the optimal road-building scheme. Note if there is no feasible solution, just output -1.
Sample Input 4 1 1 12750 28546 15361 32055 6706 3887 10754 8166 12668 19380 15788 16059 3 4 2 3 Sample Output 53246 Source
POJ Monthly--2006.01.22,zhucheng
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题意：
有 N 个牛栏，如今通过一条通道（s1,s2）把他们连起来，他们之间有一些约束关系，一些牛栏不能连在同一个点，一些牛栏必须连在同一个点，如今问有没有可能把他们都连好，并且满足全部的约束关系，假设能够，输出两个牛栏之间距离最大值的最小情况。
思路：
二分枚举最长距离。用2SAT推断可行与否。最后输出答案，假设没有，那么输出-1
条件1 i，j 相互讨厌， <i,j+n> <i+n,j> <j,i+n> <j+n,i>
条件2 i，j 关系好 <i,j> <j,i> <j+n,i+n> <i+n,j+n>
条件3
1:dis(i,s1) + dis(j,s1)>m <i,j+n> <j,i+n>
2:i j都连s2的时候与上面类似
3:dis(i,s1)+dis(s1,s2)+dis(s2,j)>m <i,j> <j+n,i+n>
4:i连s2 j连s1条件与上面类似
代码：
#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f
#define maxn 1005
#define MAXN 4000005
using namespace std;
int n,m1,m2,num,flag,ans,tot;
int head[maxn],X[2005],Y[2005],dist1[maxn],dist2[maxn];
int scc[maxn];
int vis[maxn];
int stack1[maxn];
int stack2[maxn];
struct edge
{
int v,next;
} g[MAXN];
void init()
{
memset(head,0,sizeof(head));
memset(vis,0,sizeof(vis));
memset(scc,0,sizeof(scc));
stack1[0] = stack2[0] = num = 0;
flag = 1;
}
void addedge(int u,int v)
{
num++;
g[num].v = v;
g[num].next = head[u];
head[u] = num;
}
int abs(int x)
{
if(x>=0) return x;
return -x;
}
int caldist(int x1,int y1,int x2,int y2)
{
return abs(x1-x2)+abs(y1-y2);
}
void dfs(int cur,int &sig,int &cnt)
{
if(!flag) return;
vis[cur] = ++sig;
stack1[++stack1[0]] = cur;
stack2[++stack2[0]] = cur;
for(int i = head[cur]; i; i = g[i].next)
{
if(!vis[g[i].v]) dfs(g[i].v,sig,cnt);
else
{
if(!scc[g[i].v])
{
while(vis[stack2[stack2[0]]] > vis[g[i].v])
stack2[0] --;
}
}
}
if(stack2[stack2[0]] == cur)
{
stack2[0] --;
++cnt;
do
{
scc[stack1[stack1[0]]] = cnt;
int tmp = stack1[stack1[0]];
if((tmp >= n && scc[tmp - n] == cnt) || (tmp < n && scc[tmp + n] == cnt))
{
flag = false;
return;
}
}
while(stack1[stack1[0] --] != cur);
}
}
void Twosat()
{
int i,sig,cnt;
sig = cnt = 0;
for(i=0; i<n+n&&flag; i++)
{
if(!vis[i]) dfs(i,sig,cnt);
}
}
void solve()
{
int i,j,u,v,t,le=0,ri=4000000,mid;
ans=-1;
while(le<=ri)
{
mid=(le+ri)>>1;
init();
num=0;
for(i=1;i<=m1;i++)
{
u=X[i],v=Y[i];
addedge(u,v+n);
addedge(u+n,v);
addedge(v,u+n);
addedge(v+n,u);
}
for(i=m1+1;i<=m1+m2;i++)
{
u=X[i],v=Y[i];
addedge(u,v);
addedge(v,u);
addedge(u+n,v+n);
addedge(v+n,u+n);
}
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(i==j) continue ;
if(dist1[i]+dist1[j]>mid) addedge(i,j+n);
if(dist2[i]+dist2[j]>mid) addedge(i+n,j);
if(dist1[i]+dist2[j]+tot>mid) addedge(i,j);
if(dist2[i]+dist1[j]+tot>mid) addedge(i+n,j+n);
}
}
Twosat();
if(flag)
{
ans=mid;
ri=mid-1;
}
else le=mid+1;
}
}
int main()
{
int i,j,t,x,y,x1,y1,x2,y2;
while(~scanf("%d%d%d",&n,&m1,&m2))
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
tot=caldist(x1,y1,x2,y2);
for(i=0; i<n; i++)
{
scanf("%d%d",&x,&y);
dist1[i]=caldist(x,y,x1,y1);
dist2[i]=caldist(x,y,x2,y2);
}
for(i=1;i<=m1+m2;i++)
{
scanf("%d%d",&X[i],&Y[i]);
X[i]--; Y[i]--;
}
solve();
printf("%d\n",ans);
}
return 0;
}