一个很玄乎的问题,但听到2-SAT之后就豁然开朗了。题目的意思是这样的,给你n个点群,每个点群里面有两个点,你要在每个点群里面选一个点,以这些点做半径为r的圆,然后r会有一个最大值,问的就是怎么选这些点使得r最大。
2-SAT就是对于每个变量有一些制约的关系 a->b 表示选了a就就要选b。然后我们二分这个半径,对于两点间距离<2*r的点(a,b)选了a就不能选b,选了b就不能选a,以此构图。然后跑一次强连通分量。最后判是否有解的时候就是判对于两个属于相同点群的点,它们不能处于同一强连通分量下。写的时候跪的点实在太多了,数组越界呀,强连通写错呀,精度呀,这样的题太坑爹了- -0
#pragma warning(disable:4996) #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> #include<cmath> #include<vector> #define ll long long #define maxn 220 #define eps 1e-8 using namespace std; struct Point { double x, y, z; Point(double xi, double yi, double zi) :x(xi), y(yi), z(zi){} Point(){} }p[maxn * 2]; double dist(Point a, Point b){ return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y) + (a.z - b.z)*(a.z - b.z)); } double dis[maxn * 2][maxn * 2]; int low[maxn * 2]; int pre[maxn * 2]; int dfs_clock; int sta[maxn * 2]; int st; int sccno[maxn * 2]; int n; vector<int> G[maxn * 2]; int scc_cnt; int dcmp(double x){ return (x > eps) - (x < -eps); } void dfs(int u){ low[u] = pre[u] = ++dfs_clock; sta[++st] = u; for (int i = 0; i < G[u].size(); i++){ int v = G[u][i]; if (!pre[v]){ dfs(v); low[u] = min(low[u], low[v]); } else if (!sccno[v]){ low[u] = min(low[u], pre[v]); } } if (low[u] == pre[u]){ ++scc_cnt; while (1){ int x = sta[st]; st--; sccno[x] = scc_cnt; if (x == u) break; } } } bool judge(double x) { memset(sccno, 0, sizeof(sccno)); memset(pre, 0, sizeof(pre)); memset(low, 0, sizeof(low)); st = 0; dfs_clock = 0; scc_cnt = 0; 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++){ if (dcmp(dis[i][j] - 2 * x) < 0){ G[i].push_back(j + n); G[j].push_back(i + n); } if (dcmp(dis[i][j + n] - 2 * x) < 0){ G[i].push_back(j); G[j + n].push_back(i + n); } if (dcmp(dis[i + n][j] - 2 * x) < 0){ G[i + n].push_back(j + n); G[j].push_back(i); } if (dcmp(dis[i + n][j + n] - 2 * x) < 0){ G[i + n].push_back(j); G[j + n].push_back(i); } } } for (int i = 0; i < 2 * n; i++){ if (!pre[i]) dfs(i); } for (int i = 0; i < n; i++){ if (sccno[i] == sccno[i + n]) return false; } return true; } int main() { while (cin >> n) { for (int i = 0; i < n; i++){ scanf("%lf%lf%lf", &p[i].x, &p[i].y, &p[i].z); scanf("%lf%lf%lf", &p[i + n].x, &p[i + n].y, &p[i + n].z); } for (int i = 0; i < 2 * n; i++){ for (int j = i + 1; j < 2 * n; j++){ dis[i][j] = dis[j][i] = dist(p[i], p[j]); } } double l = 0, r = 1e10; while (dcmp(r - l)>0){ double mid = (l + r) / 2; if (judge(mid)) l = mid; else r = mid; } int tmp = l * 1000; double ans = tmp / 1000.0; printf("%.3lf ", ans); } return 0; }