Bomb Game HDU - 3622

原题链接

• 题解：注意建立边
• 代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;
const int N = 5e3 + 9;
const int M = N*2*N;
int h[N], ne[M], to[M], idx;
void add(int u, int v) {
    ne[idx] = h[u], to[idx] = v, h[u] = idx++;
}
int dfn[N], low[N], times;
int sz[N], id[N], scc_cnt;
int stk[N], instk[N], top;
void tarjan(int u) {
    dfn[u] = low[u] = times++;
    stk[++top] = u;
    instk[u]= 1;
    for (int i = h[u]; ~i; i = ne[i]) {
        int v = to[i];
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if (instk[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (low[u] == dfn[u]){
        scc_cnt++;
        while (1) {
            int v = stk[top];
            top--;
            instk[v] = 0;
            id[v] = scc_cnt;
            sz[scc_cnt] ++;
            if (v == u)break;
        }
    }
}
int n, m;
struct Point {
    double x, y;
    Point() : x(), y() {}
    Point (double x, double y):x(x), y(y){}
    Point operator-(Point a) {
        return Point(x - a.x, y - a.y);
    }
    double dis() {
        return (x*x + y*y);
    }
}p[3][N];
bool check(double mid) {
    memset(h, -1, sizeof h);
    idx = 0;
    memset(dfn, 0, sizeof dfn);
    scc_cnt = 0;
    times = 0;
    for (int i = 1; i <= n; i ++) {
        for (int j = 1; j <= n; j ++) {
            if (i == j)continue;
            for (int k = 0; k < 2; k ++) {
                for (int kk = 0; kk < 2; kk ++) {
                    Point v1 = p[k][i] - p[kk][j];
                    if (v1.dis() < mid * mid) { 
                       add(i +k*n, j+ (kk^1) * n);//建边的时候要注意，必须是如果当前不满足，那么就连对面。
                       
                    }
                }
            }
        }
    }
    for (int i = 1; i <=2* n; i ++) {
        if (!dfn[i]) {
            tarjan(i);
        }
    }
    for (int i = 1; i <= n; i ++) {
        if (id[i] == id[i + n]) {
            return 0;
        }
    }
    return 1;
}
void solve() {
    for (int i = 1; i <= n; i ++) {
        scanf("%lf%lf%lf%lf",&p[0][i].x, &p[0][i].y, &p[1][i].x, &p[1][i].y);
    }
    double l = 0, r = 20000;
    int TT = 40;
    while (l < r && TT--) {
        double mid = (l + r)/2;
        if (check(mid)) {
            l = mid + 0.0000001;
        } else r = mid;
    }
    printf("%.2f\n", l*0.5);
}
int main() {
    while (~scanf("%d", &n)) {
        solve();
    }
}