POJ 2296Map Labeler 2-sat + 矩阵相交

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题目：

http://poj.org/problem?id=2296

题意：

有n个点，要用n个正方形标签贴住n个点，要求点必须在标签的上边或下边的中点，且标签不能相互覆盖，问标签的最大边长为多少

思路：

枚举两点，点在边上的位置共有上上，上下，下上，下下这四种情况，然后去判断每种情况下两个标签是否相交，即判断矩形相交，若相交，则不能共存

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int N = 210;struct edge{    int to, next;} g[N*N*10];int cnt, head[N];int num, idx, top, scc[N], st[N], dfn[N], low[N];bool vis[N];int x[N], y[N];int n, m;void add_edge(int v, int u){    g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;}void init(){    memset(head, -1, sizeof head);    memset(dfn, -1, sizeof dfn);    memset(vis, 0, sizeof vis);    cnt = num = idx = top = 0;}void tarjan(int v){    dfn[v] = low[v] = ++idx;    vis[v] = true, st[top++] = v;    int u;    for(int i = head[v]; i != -1; i = g[i].next)    {        u = g[i].to;        if(dfn[u] == -1)        {            tarjan(u);            low[v] = min(low[v], low[u]);        }        else if(vis[u]) low[v] = min(low[v], dfn[u]);    }    if(dfn[v] == low[v])    {        num++;        do        {            u = st[--top], vis[u] = false, scc[u] = num;        }        while(u != v);    }}bool judge(double axmin, double axmax, int aymin, int aymax, double bxmin, double bxmax, int bymin, int bymax){    if(axmax <= bxmin) return false;    if(aymax <= bymin) return false;    if(bxmax <= axmin) return false;    if(bymax <= aymin) return false;    return true;}bool work(int mid){    init();    for(int i = 1; i <= n; i++)        for(int j = i + 1; j <= n; j++)        {            if(judge(x[i]-mid/2.0, x[i]+mid/2.0, y[i]-mid, y[i], x[j]-mid/2.0, x[j]+mid/2.0, y[j]-mid, y[j]))                add_edge(i, j + n), add_edge(j, i + n);            if(judge(x[i]-mid/2.0, x[i]+mid/2.0, y[i]-mid, y[i], x[j]-mid/2.0, x[j]+mid/2.0, y[j], y[j]+mid))                add_edge(i, j), add_edge(j + n, i + n);            if(judge(x[i]-mid/2.0, x[i]+mid/2.0, y[i], y[i]+mid, x[j]-mid/2.0, x[j]+mid/2.0, y[j]-mid, y[j]))                add_edge(i + n, j + n), add_edge(j, i);            if(judge(x[i]-mid/2.0, x[i]+mid/2.0, y[i], y[i]+mid, x[j]-mid/2.0, x[j]+mid/2.0, y[j], y[j]+mid))                add_edge(i + n, j), add_edge(j + n, i);        }    for(int i = 1; i <= 2*n; i++)        if(dfn[i] == -1) tarjan(i);    for(int i = 1; i <= n; i++)        if(scc[i] == scc[i+n]) return false;    return true;}int main(){    int t;    scanf("%d", &t);    while(t--)    {        scanf("%d", &n);        for(int i = 1; i <= n; i++) scanf("%d%d", &x[i], &y[i]);        int l = 0, r = 20000, res;        while(l <= r)        {            int mid = (l + r) >> 1;            if(work(mid)) l = mid + 1, res = mid;            else r = mid - 1;        }        printf("%d\n", res);    }    return 0;}

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