Codeforces 848B Rooter's Song(分類+模擬)
阿新 • • 發佈:2017-09-09
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題目鏈接 Rooter‘s Song
題意 有n個舞者站在x軸上或y軸上,每個人有不同的出發時間。x軸上的舞者垂直x軸正方向移動,y軸上的舞者垂直y軸正方向移動。
當x軸的舞者和y軸的舞者相遇時,他們會互換運動軌跡。求每個舞者的最後位置。
把所有會發生碰撞的舞者塞到一起,按照坐標大小升序排序。
對於某個在x軸上的舞者, 計算他右邊的舞者個數,和他上面的舞者個數(即y軸會和他碰撞的所有舞者個數)
對於某個在y軸上的舞者, 計算他上面的舞者個數,和他右邊的舞者個數(即x軸會和他碰撞的所有舞者個數)
然後根據這些信息計算出每個舞者的碰撞次數,再求出每個舞者最後的位置。
時間復雜度$O(Mlogn)$ $(M = max(w, h))$
#include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) typedef long long LL; const int N = 4e5 + 10; const int d = 1e5; struct node{ int pos, id; friend bool operator < (const node &a, const node &b){ return a.pos < b.pos; } }; int n, w, h, op, p, t, mx, s; vector <node> X[N], Y[N]; int flag[N], ans[N], b[N], c[N]; int main(){ scanf("%d%d%d", &n, &w, &h); mx = 0; rep(i, 1, n){ scanf("%d%d%d", &op, &p, &t); b[i] = op, c[i] = p; if (op == 1) X[p - t + d].push_back({p, i}); else Y[p - t + d].push_back({p, i}); mx = max(mx, p - t + d); } rep(i, 0, mx){ sort(X[i].begin(), X[i].end()); sort(Y[i].begin(), Y[i].end()); } rep(i, 0, mx){ s = (int)X[i].size(); t = (int)Y[i].size(); if (s == 0 || t == 0) continue; rep(j, 0, s - 1){ int nx = s - j - 1, ny = t; int num; if (nx < ny) num = 2 * nx + 1; else num = ny * 2; if (num & 1) flag[X[i][j].id] = 2; else flag[X[i][j].id] = 1; if (flag[X[i][j].id] == 1) ans[X[i][j].id] = X[i][j + num / 2].pos; else ans[X[i][j].id] = Y[i][num / 2].pos; } rep(j, 0, t - 1){ int nx = s, ny = t - j - 1; int num; if (ny < nx) num = 2 * ny + 1; else num = nx * 2; if (num & 1) flag[Y[i][j].id] = 1; else flag[Y[i][j].id] = 2; if (flag[Y[i][j].id] == 2) ans[Y[i][j].id] = Y[i][j + num / 2].pos; else ans[Y[i][j].id] = X[i][num / 2].pos; } } rep(i, 1, n) if (!flag[i]){ flag[i] = b[i]; ans[i] = c[i];} rep(i, 1, n) if (flag[i] == 1) printf("%d %d\n", ans[i], h); else printf("%d %d\n", w, ans[i]); return 0; }
Codeforces 848B Rooter's Song(分類+模擬)