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): 2951    Accepted Submission(s): 984

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

 

Source

2010 Asia Regional Tianjin Site —— Online Contest

 

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解题：对半径进行二分，用2-SAT看能不能找出一个可行方案.

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
const int MAXN = 205;
const int MAXM = 40005;
const double eps = 1e-5;
int n;
int head[MAXN];
struct node1{
    int x,y;
}point[MAXN];
struct node {
    int t,next;
}edge[MAXM];
int cnt;
void add(int u, int v) {
    edge[cnt].t = v;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}
int bfn[MAXN];
int low[MAXN];
int vis[MAXN];
int s[MAXN];
int sn;
void tarbfs(int u, int lay, int & scc_num) {
    vis[u] = 1;
    bfn[u] = low[u] = lay;
    s[sn++] = u;
    int i;
    for (i = head[u]; i != -1; i = edge[i].next) {
        int tmp = edge[i].t;
        if (!vis[tmp])tarbfs(tmp, ++lay, scc_num);
        if (vis[tmp] == 1)low[u] = min(low[u], low[tmp]);
    }
    if (low[u] == bfn[u]) {
        scc_num++;
        while (1) {
            sn--;
            vis[s[sn]] = 2;
            low[s[sn]] = scc_num;
            if (s[sn] == u)break;
        }
    }
}
int tarjan() {
    int scc_num = 0;
    int lay = 1;
    int i;
    sn = 0;
    memset(vis, 0, sizeof(vis));
    for (i = 0; i < 2 * n; i++) {
        if (!vis[i])
            tarbfs(i, lay, scc_num);
    }
    return scc_num;
}
double dist(node1 a, node1 b) {
    return sqrt((double)(a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
bool solve() {
    tarjan();
    int i;
    for (i = 0; i < n/2; i++)  {
        if (low[i<<1] == low[i<<1|1])
            return false;
    }
    return true;
}
void build(double mid) {
    int i,j;
    cnt = 0;
    memset(head, -1, sizeof(head));
    for (i = 0; i < n - 2; i++) {
        int t = i;
        if (i & 1)
            t += 1;
        else
            t += 2;
        for (j = t; j < n; j++) {
            if (dist(point[i], point[j]) < 2 * mid) {
                add(i, j^1);
                add(j, i^1);
            }
        }
    }
}

int main() {
//    freopen("in.txt","r",stdin);
    int i;
    while (scanf("%d",&n) != EOF) {
        for(i = 0; i < n; i++) {
            scanf("%d %d %d %d",&point[i<<1].x, &point[i<<1].y, &point[i<<1|1].x, &point[i<<1|1].y);
        }
        n = 2 * n;
        double L = 0;
        double R = 40000;
        while (R - L > eps) {
            double mid = (L + R) / 2;
            build(mid);
            if (solve())
                L = mid;
            else
                R = mid;
        }
        printf("%.2lf\n",L);
    }
    return 0;
}