poj 2749 Building roads （二分+拆點+2-sat）

Building roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6229   Accepted: 2093 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 112750 28546 15361 320556706 388710754 816612668 1938015788 160593 42 3 Sample Output 53246 Source POJ Monthly--2006.01.22,zhucheng

題意：

有 N 個牛欄，現在通過一條通道（s1,s2）把他們連起來，他們之間有一些約束關係，一些牛欄不能連在同一個點，一些牛欄必須連在同一個點，現在問有沒有可能把他們都連好，而且滿足所有的約束關係，如果可以，輸出兩個牛欄之間距離最大值的最小情況。

思路：

二分枚舉最長距離。用2SAT判斷可行與否。最後輸出答案，如果沒有，那麼輸出-1
條件1   i，j  相互討厭，  <i,j+n>  <i+n,j>  <j,i+n>  <j+n,i>
條件2   i，j   關係好       <i,j> <j,i> <j+n,i+n> <i+n,j+n>
條件3
     1:dis(i,s1) + dis(j,s1)>m     <i,j+n>  <j,i+n>
     2:i j都連s2的時候與上面類似
     3:dis(i,s1)+dis(s1,s2)+dis(s2,j)>m   <i,j>  <j+n,i+n>
     4:i連s2 j連s1條件與上面類似

代碼：

#include <cstdio>#include <cstring>#define INF 0x3f3f3f3f#define maxn 1005#define MAXN 4000005using 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;}