poj 2749(二分+2-SAT)

poj 2749
1、加边视情况加
2、2-SAT里面的c在solve里面千万不要再int c

#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm>
#define ll long long
#define db double
#define ms(i,j) memset(i,j)
#define INF 100000000
#define llINF 1LL<<60
using namespace std;

const int MAXN = (50000 + 5) * 2, LEFT = 0, RIGHT = 1;

struct TwoSAT
{
    vector<int> G[MAXN*2];
    int mark[MAXN*2];
    int S[MAXN*2];
    int n, c;
    void init(int n)
    {
        this->n = n;
        for (int i=0;i<=n*2;i++)
        {
            mark[i] = false;
            G[i].clear();
        }
    }
    void addE(int x, int xv, int y, int yv)
    {
        x = x*2 + xv;
        y = y*2 + yv;
        G[x^1].push_back(y);
        G[y^1].push_back(x);
    }
    int dfs(int x)
    {
        if (mark[x^1]) return false;
        if (mark[x]) return true;
        mark[x] = true;
        S[c++] = x;
        for (int i=0;i<G[x].size();i++)
        {
            if (!dfs(G[x][i])) return false;
        }
        return true;
    }
    bool solve()
    {
        for (int i = 0;i < n;i += 2)
        if (!mark[i]&&!mark[i+1])
        {
            c = 0;
            if (!dfs(i))
            {
                for (int j = 0; j < c; j++)
                        mark[S[j]] = false;
                    c = 0;
                if (!dfs(i+1)) return false;
            }
        }
        return true;
    }
}ts;

int n, A, B, sx1, sy1, sx2, sy2, xi[MAXN], yi[MAXN], hate[MAXN][2], fri[MAXN][2], d1[MAXN], d2[MAXN],di;

int dis(int x1, int y1, int x2, int y2)
{
    return abs(x1-x2)+abs(y1-y2);
}
int judge(int mid)
{
    ts.init(n);
    for (int i=0;i<n;i++)
    for (int j=i+1;j<n;j++)
    {
        int x = i, y = j;
        if (d1[i]+d1[j]>mid)     ts.addE(x, 1, y, 1);
        if (d2[i]+d2[j]>mid)    ts.addE(x, 0, y, 0);
        if (d1[i]+di+d2[j]>mid)    ts.addE(x, 1, y, 0);
        if (d2[i]+di+d1[j]>mid) ts.addE(x, 0, y, 1);
    }
    for (int i=0;i<A;i++)
    {
        int x = hate[i][LEFT] - 1, y = hate[i][RIGHT] - 1;
        ts.addE(x, 0, y, 0);
        ts.addE(x, 1, y, 1);
    }
    for (int i=0;i<B;i++)
    {
        int x = fri[i][LEFT] - 1, y = fri[i][RIGHT] - 1;
        ts.addE(x, 0, y, 1);
        ts.addE(y, 0, x, 1);
    }
    return ts.solve();
}
int init()
{
    scanf("%d%d%d%d", &sx1, &sy1, &sx2, &sy2);
    for (int i=0;i<n;i++)
    {
        scanf("%d%d", &xi[i], &yi[i]);
    }
    for (int i=0;i<A;i++)
    {
        scanf("%d%d", &hate[i][LEFT], &hate[i][RIGHT]);
    }
    for (int i=0;i<B;i++)
    {
        scanf("%d%d", &fri[i][LEFT], &fri[i][RIGHT]);
    }
}
int solve()
{
    di = dis(sx1, sy1, sx2, sy2);
    for (int i=0;i<n;i++)
    {
        d1[i] = dis(xi[i], yi[i], sx1, sy1);
        d2[i] = dis(xi[i], yi[i], sx2, sy2);
    }
    int l = 0, r = 8000000, ans = 10000000;
    while (l < r)
    {
        int mid = (l + r) / 2;
        if (judge(mid))
        {
            ans = mid;
            r = mid;
        } else l = mid+1;
    }
    if (ans!=10000000)
    {
        printf("%d\n", ans);
    } else printf("%d\n", -1);
}
int main()
{
    while (scanf("%d%d%d", &n, &A, &B)==3)
    {
        init();
        solve();
    }
    return 0;
}