POJ2749 Building roads【2-SAT】

题意：有很多仓库，必须连接两个中转站中的一个（S1，S2），以至于可以与任意的仓库连接。有些仓库不能连相同的点，有些必须连相同的点，问所有方案中，最大距离的最小值

思路：二分最大距离，枚举每两个仓库，如果它们的某一种连接方法距离 大于 二分的距离的话，这种连接法矛盾，建边，SCC是否可行。一开始来一次SCC，来判断是否有方案。

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<stdlib.h>#include<math.h>#include<vector>#include<list>#include<map>#include<stack>#include<queue>#include<algorithm>#include<numeric>#include<functional>using namespace std;typedef long long ll;const int maxn = 5005;const int maxm = 8e6+5;struct edge{int to,next;}ed[maxm];struct data{int x,y;}node[maxn],s1,s2,bian[maxn];int head[maxn], tot; int Low[maxn], DFN[maxn], Stack[maxn];  int Index, top;bool Instack[maxn]; int block,belong[maxn];  int zhong,cnt;void init(){memset(head, -1, sizeof(head));      memset(DFN, 0, sizeof(DFN));       memset(Instack, false, sizeof(Instack));       block = tot = 0;    Index = top = 0;}void Tarjan(int u,int pre){    int v;    Low[u] = DFN[u] = ++Index;    Stack[top++] = u;    Instack[u] = true;    for(int i = head[u];i != -1;i = ed[i].next)    {        v = ed[i].to;        if( !DFN[v] )        {            Tarjan(v,u);            if( Low[u] > Low[v] )Low[u] = Low[v];        }        else if( Instack[v] && Low[u] > DFN[v] )            Low[u] = DFN[v];    }    if(Low[u] == DFN[u])    {        block++;        do        {            v = Stack[--top];            Instack[v] = false;            belong[v] = block;        }        while( v!=u );    }}void add(int x,int y){ed[tot].to = y;ed[tot].next = head[x];head[x] = tot++;}void SCC(int n){      for(int i = 1; i <= n; i++)          if(!DFN[i])             Tarjan(i,-1);  }int cal(int a,int b){if(b == 1)return abs(node[a].x-s1.x) + abs(node[a].y-s1.y);elsereturn abs(node[a].x-s2.x) + abs(node[a].y-s2.y);}int check(int dis,int n){init();for(int i = 0; i < cnt; i++)add(bian[i].x,bian[i].y);for(int i = 1; i <= n; i++){for(int j = i+1; j <= n; j++){if(cal(i,1) + cal(j,2)+zhong > dis){add(i,j);add(j+n,i+n);}if(cal(i,2) + cal(j,1)+zhong > dis){add(i+n,j+n);add(j,i);}if(cal(i,1) + cal(j,1) > dis){add(i,j+n);add(j,i+n);}if(cal(i,2) + cal(j,2) > dis){add(i+n,j);add(j+n,i);}}}SCC(2*n);int flag = 1;for(int i = 1; i <= n; i++){if(belong[i] == belong[i+n]){flag = 0;break;}}return flag;}int main(void){int n,m,a,b,c,d;while(scanf("%d%d%d",&n,&c,&d)!=EOF && n+m){init();scanf("%d%d%d%d",&s1.x,&s1.y,&s2.x,&s2.y);for(int i = 1; i <= n; i++)scanf("%d%d",&node[i].x,&node[i].y);cnt = 0;while(c--){scanf("%d%d",&a,&b);add(a,b+n);add(b,a+n);add(a+n,b);add(b+n,a);bian[cnt].x = a;bian[cnt++].y = b+n;bian[cnt].x = b;bian[cnt++].y = a+n;bian[cnt].x = a+n;bian[cnt++].y = b;bian[cnt].x = b+n;bian[cnt++].y = a;}while(d--){scanf("%d%d",&a,&b);add(a,b);add(b+n,a+n);add(a+n,b+n);add(b,a);bian[cnt].x = a;bian[cnt++].y = b;bian[cnt].x = b+n;bian[cnt++].y = a+n;bian[cnt].x = a+n;bian[cnt++].y = b+n;bian[cnt].x = b;bian[cnt++].y = a;}SCC(2*n);int flag = 1;for(int i = 1; i <= n; i++){if(belong[i] == belong[i+n]){flag = 0;break;}}if(!flag){printf("-1\n");continue;}else{zhong = abs(s1.x - s2.x) + abs(s1.y - s2.y);int l = 0,r = 9e6+5,mid;while(l < r){int mid = (l+r) / 2;if(check(mid,n))r = mid;elsel = mid + 1;}printf("%d\n",l);}}return 0;}