HDU 3622 Bomb Game（二分+2-SAT）

Bomb Game

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5396    Accepted Submission(s): 1925

Problem Description

Robbie is playing an interesting computer game. The game field is an unbounded 2-dimensional region. There are N rounds in the game. At each round, the computer will give Robbie two places, and Robbie should choose one of them to put a bomb. The explosion area of the bomb is a circle whose center is just the chosen place. Robbie can control the power of the bomb, that is, he can control the radius of each circle. A strange requirement is that there should be no common area for any two circles. The final score is the minimum radius of all the N circles.
Robbie has cracked the game, and he has known all the candidate places of each round before the game starts. Now he wants to know the maximum score he can get with the optimal strategy.

 

Input

The first line of each test case is an integer N (2 <= N <= 100), indicating the number of rounds. Then N lines follow. The i-th line contains four integers x_1i, y_1i, x_2i, y_2i, indicating that the coordinates of the two candidate places of the i-th round are (x_1i, y_1i) and (x_2i, y_2i). All the coordinates are in the range [-10000, 10000].

 

Output

Output one float number for each test case, indicating the best possible score. The result should be rounded to two decimal places.

 

Sample Input

2

1 1 1 -1

-1 -1 -1 1

2

1 1 -1 -1

1 -1 -1 1

 

Sample Output

1.41

1.00

 

 

题目链接：HDU 3622

很明显的二分题，二分爆炸半径r使得被选中的炸弹的爆炸出的圆面积均没有相交（边缘可以重叠）即可。

代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const double eps = 1e-5;
const int N = 110;
const int M = N * N * 4;
struct edge
{
    int to, nxt;
    edge() {}
    edge(int _to, int _nxt): to(_to), nxt(_nxt) {}
};
pii P[N];
edge E[M];
int head[N << 1], tot;
int dfn[N << 1], low[N << 1], st[N << 1], belong[N << 1], ts, scc, top;
bitset < N << 1 > ins;
int n;

void init()
{
    CLR(head, -1);
    tot = 0;
    CLR(dfn, 0);
    CLR(low, 0);
    CLR(belong, 0);
    ts = scc = top = 0;
    ins.reset();
}
inline void add(int s, int t)
{
    E[tot] = edge(t, head[s]);
    head[s] = tot++;
}
void Tarjan(int u)
{
    dfn[u] = low[u] = ++ts;
    st[top++] = u;
    ins[u] = 1;
    int i, v;
    for (i = head[u]; ~i; i = E[i].nxt)
    {
        v = E[i].to;
        if (!dfn[v])
        {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if (ins[v])

            low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u])
    {
        ++scc;
        do
        {
            v = st[--top];
            ins[v] = 0;
            belong[v] = scc;
        }
        while (u != v);
    }
}
inline double cal(const pii &a, const pii &b)
{
    return sqrt((a.first - b.first) * (a.first - b.first) + (a.second - b.second) * (a.second - b.second));
}
bool check(double r, int sz)
{
    init();
    int i, j;
    for (i = 0; i < sz; i += 2)
    {
        for (j = i + 2; j < sz; ++j)
        {
            double dx = cal(P[i], P[j]);
            if (dx < 2 * r)
            {
                add(i, j ^ 1);
                add(j, i ^ 1);
            }
        }
    }
    for (i = 1; i < sz; i += 2)
    {
        for (j = i + 1; j < sz; ++j)
        {
            double dx = cal(P[i], P[j]);
            if (dx < 2 * r)
            {
                add(i, j ^ 1);
                add(j, i ^ 1);
            }
        }
    }
    for (i = 0; i < sz; ++i)
        if (!dfn[i])
            Tarjan(i);
    for (i = 0; i < sz; ++i)
        if (belong[i] == belong[i ^ 1])
            return false;
    return true;
}
int main(void)
{
    int i;
    while (~scanf("%d", &n))
    {
        for (i = 0; i < (n << 1); i += 2)
            scanf("%d%d%d%d", &P[i].first, &P[i].second, &P[i | 1].first, &P[i | 1].second);
        double L = 0, R = 10000 * 1.45;
        double ans = 0;
        while (R - L >= eps)
        {
            double mid = (L + R) / 2.0;
            if (check(mid, n << 1))
            {
                L = mid;
                ans = mid;
            }
            else
                R = mid;
        }
        printf("%.2f
", ans);
    }
    return 0;
}