HDU 3622 Bomb Game（二分 + 2-SAT）

点击打开链接

题意：给出n对点，每对点只能选取一个。每个点都会以自己为中心进行爆炸且半径相同，问最大爆炸半径是多少。

解法：两个之中选一个，可以考虑2-SAT求解。由于半径是要求的，所以二分半径，然后每次都建图用2-SAT判断看是否可行。

2-SAT建图中的边代表的是，如果选取i，就一定要选取j。对于这道题，如果有两个点的距离小于爆炸距离，即不符合，那么认为选取了i 只能选取 j^1。选取了j只能选取 i^1。

代码如下：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<utility>
#include<stack>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<map>
using namespace std;
const int maxn = 1005 * 2;
const int maxm = 1e6 + 5;
const double eps = 1e-4;
stack <int> sta;
int dfn[maxn], low[maxn], key[maxn], tot = 0, num = 1;
int belong[maxn];
int n, m;
int a[maxn], b[maxn];

struct Node {
	int x1, x2, y1, y2;
}boom[maxn];

int to[maxm], nx[maxm], head[maxn], ppp;
void add_edge(int u, int v) {
	to[ppp] = v, nx[ppp] = head[u], head[u] = ppp++;
}

void tarjan(int u) {
	sta.push(u);
	dfn[u] = low[u] = tot++;
	key[u] = 1;
	for(int i = head[u]; ~i; i = nx[i]) {
		int v = to[i];
		if(dfn[v] == -1) {
			tarjan(v);
			low[u] = min(low[u], low[v]);
		} else if(key[v] == 1) {
			low[u] = min(low[u], dfn[v]);
		}
	}
	if(dfn[u] == low[u]) {
		while(1) {
			int v = sta.top();
			sta.pop();
			belong[v] = num;
			key[v] = 2;
			if(v == u)
				break;
		}
		num++;
	}
}

double Count(int x1, int y1, int x2, int y2) {
	return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

void build(double mid) {
	mid *= 2.0;
	for(int i = 1, a, b; i <= n; i++) {
		for(int j = i + 1; j <= n; j++) {
			double tmp1 = Count(boom[i].x1, boom[i].y1, boom[j].x1, boom[j].y1);
			double tmp2 = Count(boom[i].x1, boom[i].y1, boom[j].x2, boom[j].y2);
			double tmp3 = Count(boom[i].x2, boom[i].y2, boom[j].x1, boom[j].y1);
			double tmp4 = Count(boom[i].x2, boom[i].y2, boom[j].x2, boom[j].y2);
			if(tmp1 < mid) {
				a = i * 2, b = j * 2;
				add_edge(a, b ^ 1);
				add_edge(b, a ^ 1);
			}
			if(tmp4 < mid) {
				a = i * 2 + 1, b = j * 2 + 1;
				add_edge(a, b ^ 1);
				add_edge(b, a ^ 1);
			}
			if(tmp2 < mid) {
				a = i * 2, b = j * 2 + 1;
				add_edge(a, b ^ 1);
				add_edge(b, a ^ 1);
			}
			if(tmp3 < mid) {
				a = i * 2 + 1, b = j * 2;
				add_edge(a, b ^ 1);
				add_edge(b, a ^ 1);
			}
		}
	}
}

bool Check() {
	for(int i = 1; i <= n; i++)
		if(belong[2 * i] == belong[(2 * i) ^ 1]) 
			return 0;
	return 1;
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
#endif
	while(scanf("%d", &n) != EOF) {
		for(int i = 1; i <= n; i++) {
			scanf("%d%d%d%d", &boom[i].x1, &boom[i].y1, &boom[i].x2, &boom[i].y2);
		}
		double front = 0, back = 1e9+7;
		while(back - front > eps) {
//			cout << front << ' ' << back << '\n';
			memset(head, -1, sizeof(head));
			memset(dfn, -1, sizeof(dfn));
			memset(key, 0, sizeof(key));
			ppp = tot = 0;
			num = 1;
			double mid = (front + back) / 2.0;
			build(mid);
			for(int i = 2; i <= 2 * n + 1; i++) {
				if(dfn[i] == -1)
					tarjan(i);
			}
			if(Check())
				front = mid;
			else
				back = mid;
		}
		printf("%.2f\n", front);
	}
	return 0;
}