[hihoCoder] 穿越禁区 | 隐式图两点可达性判断

http://hihocoder.com/problemset/problem/1307

首先，如果左、右边界被圆分离开，就意味着无法穿越雷区。

把上、下边界以及N个圆抽象成N+2个图节点，当边界与圆或圆与圆之间相交时表示它们连通，问题就转变为判断从上边界到下边界是否存在路径。建图后从上边界跑一遍DFS即可。

#include <iostream>
#include <cstdio>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cmath>

using namespace std;

struct Circle
{
    long long x, y, r;
}circles[1005];

int W, H, N;

vector<vector<int> > adj;

bool intersect(Circle& a, Circle& b)
{
    return ( sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) ) - (a.r+b.r) <= 0 );
}

void dfs(int cur, vector<bool>& vis)
{
    vis[cur] = true;
    for (int i = 0; i < adj[cur].size(); ++i) {
        int to = adj[cur][i];
        if (!vis[to]) {
            dfs(to, vis);
        }
    }
}

void solve()
{
    adj = vector<vector<int> >(N+2);
    int s = 0, t = N+1;
    // up (0)
    for (int i = 1; i <= N; ++i) {
        if (circles[i].y + circles[i].r >= H) {
            adj[s].push_back(i);
        }
        if (circles[i].y - circles[i].r <= 0) {
            adj[i].push_back(t);
        }
    }
    for (int i = 1; i < N; ++i) {
        for (int j = i + 1; j <= N; ++j) {
            if (intersect(circles[i], circles[j])) {
                adj[i].push_back(j);
                adj[j].push_back(i);
            }
        }
    }
    // DFS
    vector<bool> vis(N+2, false);
    dfs(s, vis);
    if (vis[t]) {
        cout << "NO" << endl;
    } else {
        cout << "YES" << endl;
    }
}

int main()
{
    int T;
    cin >> T;
    for (int i = 0; i < T; ++i) {
        cin >> W >> H >> N;
        for (int j = 1; j <= N; ++j) {
            scanf("%lld%lld%lld", &circles[j].x, &circles[j].y, &circles[j].r);
        }
        solve();
    }
    return 0;
}