HDU

题目大意：玩一个放炸弹游戏，有N次放炸弹的机会，每次放炸弹时，你都有两个位置能够选择。问怎样放炸弹，能使爆炸的炸弹的半径的最小值最大(炸弹爆炸半径能够控制，可是爆炸形成的圈不能有重叠部分)

解题思路：最小值最大，二分
二分半径，假设有不满足的点，就建立起限制边。接着推断是否能完毕染色就可以

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
#define esp 1e-5
#define N 210 

struct Node {
    int x, y;
}node[N][2];

bool mark[N];
vector<int> G[N];
int top, n, m;
int S[N];

void init (){
    for (int i = 0; i < n ; i++) {
        scanf("%d%d%d%d", &node[i][0].x, &node[i][0].y, &node[i][1].x, &node[i][1].y);
    }
}
double distance(int i, int a, int j, int b) {
    return sqrt( 1.0 * (node[i][a].x - node[j][b].x) * (node[i][a].x - node[j][b].x) + 1.0 * (node[i][a].y - node[j][b].y) * (node[i][a].y - node[j][b].y));
}

void AddEdge(int i, int a, int j, int b) {
    int x = i * 2 + a;
    int y = j * 2 + b;
    G[x ^ 1].push_back(y);
    G[y ^ 1].push_back(x);
}

bool dfs(int u) {
    if (mark[u ^ 1])
        return false;
    if (mark[u])
        return true;
    mark[u] = true;
    S[++top] = u;
    for (int i = 0; i < G[u].size(); i++)
        if (!dfs(G[u][i]))
            return false;
    return true;
}

bool judge(double mid) {
    for (int i = 0; i < 2 * n; i++)
        G[i].clear();
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            for (int a = 0; a < 2; a++)
                for (int b = 0; b < 2; b++) {
                    //printf("dis is %lf
", distance(i, a, j, b));
                    if (2 * mid - distance(i, a, j, b) > esp) {
                        AddEdge(i, a, j, b);
                    }
                }
            }
    }

    memset(mark, 0, sizeof(mark));
    for (int i = 0; i < 2 * n; i += 2) {
        if (!mark[i] && !mark[i ^ 1]) {
            top = 0;
            if (!dfs(i)) {
                while(top) mark[S[top--]] = false;
                if (!dfs(i ^ 1))
                    return false;
            }
        }
    }
    return true;
}

void solve() {
    double l = 0, r = 1e5;
    for (int i = 0; i < 30; i++) {
        double mid = (l + r) / 2;
//      printf("l is %lf r is %lf mid is %lf
", l, r, mid);
        if (judge(mid))
            l = mid;
        else
            r = mid;
    }
    printf("%.2lf
", l);
}

int main() {
    while (scanf("%d", &n) != EOF) {
        init();
        solve();
    }
    return 0;
}