题目链接

  本题是2009年icpc欧洲区域赛西北欧赛区的j题

  UVA - 12227 Wormholes

题意

  你有一艘星际飞船，飞船运行速度为1，打算从坐标a旅行到坐标b（出发时刻为0），已知n个虫洞（0≤n≤ 50)的信息：入口点坐标s，出口点坐标e，形成时间t，穿过虫洞后时间的迁移量d(−1000000≤t,d≤1000000)。所有坐标的x、y、z绝对值不超过10000，并且任意两点间的距离简化成欧式距离向上取整。求达到b点的最早时间。

分析

  由于穿过虫洞后时间的迁移量d可能为负数，一旦经过某虫洞存在负圈，那么即便首次到达虫洞入口的时间晚于其形成时间t，只要绕着此圈不停地来回，到达虫洞两端点的最早时间将回退到t和t+d。

  因此只要预处理虫洞，看每个虫洞是否能找出负圈，如果能找到负圈那么达到虫洞两端的最早时间为$t_s=min(S_{as},t)$和$min(t_s+S_{se},t+d)$。预处理之后，跑一边Bellman Ford就能得到答案。对每个虫洞预处理看能否找出负圈的过程类似Bellman Ford，不过实际上更高效，参见代码。

测试数据

  给一份测试数据

AC 代码

#include <iostream>
#include <queue>
using namespace std;#define N 110
int s[N][N], f[N], g[N], t[N], d[N], x[N], y[N], z[N], m, n; bool inq[N], cyc[N];int ceil_sqrt(int x) {int l = 0, r = min((x+1)>>1, 34642);while (l <= r) {int m = (l+r+1)>>1, y = m*m;if (y == x) return m;y < x ? l = m+1 : r = m-1;}return l;
}bool update(int i) {queue<int> q;for (int j=2; j<n; j+=2) {g[j] = (cyc[j] ? t[j] : max(t[i]+d[i]+s[i+1][j], t[j])) + d[j];if (g[j] + s[j+1][i] < t[i]) {f[i] = min(f[i], t[i]); f[i+1] = min(f[i+1], min(f[i]+s[i][i+1], t[i]+d[i]));return cyc[i] = true;}q.push(j); inq[j] = true;}while (!q.empty()) {int u = q.front(); q.pop(); inq[u] = false;for (int j=2; j<n; j+=2) if (!cyc[j]) {int v = max(g[u]+s[u+1][j], t[j]) + d[j];if (v < g[j]) {if (v + s[j+1][i] < t[i]) {f[i] = min(f[i], t[i]); f[i+1] = min(f[i+1], min(f[i]+s[i][i+1], t[i]+d[i]));return cyc[i] = true;}g[j] = v;if (!inq[j]) q.push(j), inq[j] = true;}}}return false;
}void solve() {cin >> x[0] >> y[0] >> z[0] >> x[1] >> y[1] >> z[1] >> m; n = 2;for (int i=0; i<m; ++i) cin >> x[n] >> y[n] >> z[n++] >> x[n] >> y[n] >> z[n++] >> t[n-2] >> d[n-2];for (int i=0; i<n; ++i) for (int j=i; j<n; ++j) s[i][j] = s[j][i] = ceil_sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])+(z[i]-z[j])*(z[i]-z[j]));for (int i=0; i<n; ++i) f[i] = s[i][0], cyc[i] = false;for (int i=2; i<n; i+=2) if (s[i][i+1] + d[i] < 0)f[i] = min(f[i], t[i]), f[i+1] = min(f[i+1], min(f[i]+s[i][i+1], t[i]+d[i])), cyc[i] = true;while (true) {bool updated = false;for (int i=0; i<n; i+=2) if (!cyc[i] && update(i)) updated = true;if (!updated) break;}queue<int> q;for (int i=2; i<n; ++i) q.push(i), inq[i] = true;while (!q.empty()) {int u = q.front(); q.pop(); inq[u] = false;for (int i=0, v; i<n; ++i) if ((v = s[i][u] + f[u]) < f[i]) {f[i] = v;if (!inq[i]) q.push(i), inq[i] = true;}if (u&1) continue;int w = u+1, v = max(f[u], t[u]) + d[u];if (v < f[w]) {f[w] = v;if (!inq[w]) q.push(w), inq[w] = true;}}for (int i=2; i<n; ++i) f[1] = min(f[1], f[i] + s[1][i]);cout << f[1] << endl;
}int main() {ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);int t; cin >> t;while (t--) solve();return 0;
}