Bomb Game（二分 2-SAT）

传送门

题意

略

分析

要求最大的合法半径。
可以考虑二分，每次check重新建图跑tarjan判断是否合法。
小于ans的半径一定合法，本题是有二分性质的。
（数组开小了一直T调了好久）

代码

#include <bits/stdc++.h>

using namespace std;
//-----pre_def----
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
#define fir(i, a, b) for (int i = (a); i <= (b); i++)
#define rif(i, a, b) for (int i = (a); i >= (b); i--)
#define endl '\n'
#define init_h memset(h, -1, sizeof h), idx = 0;
#define lowbit(x) x &(-x)

//---------------
const int N = 200 + 10;
const double eps = 1e-5;
double a[N], b[N], c[N], d[N];
int read()
{
    int ret = 0, flag = 0;
    char ch;
    if ((ch = getchar()) == '-')
        flag = 1;
    else if (ch >= '0' && ch <= '9')
        ret = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9')
        ret = ret * 10 + (ch - '0');
    return flag ? -ret : ret;
}
struct two_SAT
{
    int n; //未翻倍之前的点数
    bool in_stack[N];
    int id[N], stk[N], top, dfn[N], low[N], times, scc_cnt; //所在scc编号，栈，dfs序，最下能到哪个节点，时间戳，scc编号
    int h[N], e[N*N], ne[N*N], idx;//这里数组开小了，一直TLE，，，
    void init(int _n)
    {
        n = _n;
        init_h;
    }
    void over()
    {
        memset(id, 0, sizeof id);
        memset(dfn, 0, sizeof dfn);
        memset(low, 0, sizeof low);
        memset(in_stack,0,sizeof in_stack);
        times = top = scc_cnt = 0;
    }
    inline void add(int a, int b)
    {
        e[idx] = b;
        ne[idx] = h[a];
        h[a] = idx++;
    }
    inline void add_and(int x, int valx, int y, int valy) //如果x取valx，则y必须取valy
    {
        //x -> y or !x -> y or x -> !y or !x -> !y
        add(x + valx * n, y + valy * n);
    }
    void add_or(int x, int valx, int y, int valy) //x取valx 或 y取valy
    {
        add(x + (!valx) * n, y + valy * n);
        add(y + (!valy) * n, x + valx * n);
    }
    void add_one(int x, int valx)
    {
        add_or(x, valx, x, valx);
    }
    bool check(double x1, double y1, double x2, double y2, double r)
    {
        return (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) < 2 * 2 * r * r;
    }
    void build(double dis)
    {
        fir(i, 1, n)
        {
            fir(j, i + 1, n)
            {
                //i,j;
                if (check(a[i], b[i], a[j], b[j], dis))
                {
                    add_and(i, 0, j, 1);
                    add_and(j, 0, i, 1);
                }
                if (check(a[i], b[i], c[j], d[j], dis))
                {
                    add_and(i, 0, j, 0);
                    add_and(j, 1, i, 1);
                }
                if (check(c[i], d[i], a[j], b[j], dis))
                {
                    add_and(i, 1, j, 1);
                    add_and(j, 0, i, 0);
                }
                if (check(c[i], d[i], c[j], d[j], dis))
                {
                    add_and(i, 1, j, 0);
                    add_and(j, 1, i, 0);
                }
            }
        }
    }
    void tarjan(int u)
    {
        dfn[u] = low[u] = ++times;
        stk[++top] = u, in_stack[u] = true;
        for (int i = h[u]; ~i; i = ne[i])
        {
            int j = e[i];
            if (!dfn[j])
            {
                tarjan(j);
                low[u] = min(low[u], low[j]);
            }
            else if (in_stack[j])
            {
                low[u] = min(low[u], dfn[j]);
            }
        }
        if (dfn[u] == low[u])
        {
            ++scc_cnt;
            int y;
            do
            {
                y = stk[top--];
                in_stack[y] = false;
                id[y] = scc_cnt;
            } while (y != u);
        }
    }
    void tarjan()
    {
        fir(i, 1, n << 1) if (!dfn[i])
            tarjan(i);
    }
    bool solve()
    {
        tarjan();
        for (int i = 1; i <= n; i++)
            if (id[i] == id[i + n])
                return false;
        return true;
    }
} t;

void init() {}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    int StartTime = clock();
#endif
    int n;
    while (~scanf("%d", &n))
    {
        fir(i, 1, n)
        {
            scanf("%lf%lf%lf%lf", &a[i], &b[i], &c[i], &d[i]);
        }
        double l = 0, r = 40000;
        while(r-l>eps)
        {
            double mid = (l + r) / 2.0;
            t.init(n);
            t.build(mid);
            if (t.solve())
            {
                l = mid;
            }
            else
            {
                r = mid;
            }
            t.over();
        }
        printf("%.2lf\n", l);
    }
#ifndef ONLINE_JUDGE
    printf("Run_Time = %d ms\n", clock() - StartTime);
#endif
    return 0;
}