【題解】 「NOI2019」彈跳 KDT優化建圖+最短路+剪枝+卡空間 LOJ3159
阿新 • • 發佈：2020-09-27
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有 \(n\ (1 \le n \le 70000)\) 個二維平面上的點，\(m\ (1 \le m \le 150000)\) 條集體邊，每條集體邊為從某個點向一個矩形所在的所有點連邊。
求 \(1\) 到所有點的最短路。
時空限制：\(\textrm{2s/128MB}\)。
Editorial
很容易讓人想到 \(\rm{xxx}\) 優化建圖，發現非常無趣的是，二維線段樹是兩個 \(\log\)。被卡啦！
於是我們果斷用 \(\rm{kdtree}\)！發現還是被卡空間啦！
於是我們考慮不把邊建出來，而是直接對整棵子樹取 \(\min\)。
\(\rm{kdtree}\) 恰好是一棵平衡樹，也可以維護全域性最小值！
所以你就可以很方便的寫 \(\rm{Dijkstra}\) 啦！
複雜度？\(O(m \sqrt{n})\)！
Code
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\(\rm{kdtree}\) 真的難寫。
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\(\rm{kdtree}\) 不剪枝？\(\rm{TLE}\) 歡迎你。
#include <bits/stdc++.h> #define debug(...) ; // fprintf(stderr ,__VA_ARGS__) #define __FILE(x)\ freopen(#x".in" ,"r" ,stdin);\ freopen(#x".out" ,"w" ,stdout) int read(){ char k = getchar(); int x = 0; while(k < '0' || k > '9') k = getchar(); while(k >= '0' && k <= '9') x = x * 10 + k - '0' ,k = getchar(); return x; } inline void chkmin(int &x ,int y){x = std::min(x ,y);} inline void chkmax(int &x ,int y){x = std::max(x ,y);} const int MX = 2e5 + 233; struct point{ int x[2] ,w ,id; point(int X ,int Y ,int W = 0 ,int ID = 0){x[0] = X ,x[1] = Y ,w = W ,id = ID;} point(int a[2] ,int W = 0 ,int ID = 0){x[0] = a[0] ,x[1] = a[1] ,w = W ,id = ID;} point(){x[0] = x[1] = id = 0 ,w = INT_MAX;} }P[MX]; struct node{ int mn[2] ,mx[2] ,ch[2]; int del ,covmn ,mnval ,mnfr ,val ,alive; int mxval; point tp; }; int D; bool operator <(point a ,point b){return a.x[D] < b.x[D];} int trans[MX] ,tkdt[MX]; struct KDT{ #define lch tr[x].ch[0] #define rch tr[x].ch[1] node tr[MX] ; int cnt ,rt ,D; KDT(){ for(int i : {0 ,1}){ tr[0].mx[i] = INT_MIN; tr[0].mn[i] = INT_MAX; } tr[0].covmn = tr[0].mnval = tr[0].val = INT_MAX; tr[0].mnfr = tr[0].alive = false; tr[0].del = true; tr[0].mxval = INT_MIN; } void docovmn(int x ,int v){ if(!tr[x].del) chkmin(tr[x].val ,v); if(tr[x].alive){ chkmin(tr[x].mnval ,v); chkmin(tr[x].covmn ,v); chkmin(tr[x].mxval ,v); } } void _pushup(int x){ if(!tr[x].del){ tr[x].mnval = tr[x].val; tr[x].mxval = tr[x].val; tr[x].mnfr = x; } else{ tr[x].mnfr = 0 ,tr[x].mnval = INT_MAX; tr[x].mxval = INT_MIN; } tr[x].alive = tr[lch].alive || tr[rch].alive || !tr[x].del; for(int i : {0 ,1}){ tr[x].mn[i] = tr[x].mx[i] = tr[x].tp.x[i]; for(int j : {lch ,rch}){ chkmin(tr[x].mn[i] ,tr[j].mn[i]); chkmax(tr[x].mx[i] ,tr[j].mx[i]); chkmin(tr[x].mnval ,tr[j].mnval); chkmax(tr[x].mxval ,tr[j].mxval); if(tr[x].mnval == tr[j].mnval){ tr[x].mnfr = tr[j].mnfr; } } } } void pushup(int x){ if(!tr[x].del){ tr[x].mnval = tr[x].val; tr[x].mxval = tr[x].val; tr[x].mnfr = x; } else{ tr[x].mnfr = 0 ,tr[x].mnval = INT_MAX; tr[x].mxval = INT_MIN; } tr[x].alive = tr[lch].alive || tr[rch].alive || !tr[x].del; for(int j : {lch ,rch}){ chkmin(tr[x].mnval ,tr[j].mnval); chkmax(tr[x].mxval ,tr[j].mxval); if(tr[x].mnval == tr[j].mnval){ tr[x].mnfr = tr[j].mnfr; } } } void pushdown(int x){ if(tr[x].covmn != INT_MAX){ if(lch) docovmn(lch ,tr[x].covmn); if(rch) docovmn(rch ,tr[x].covmn); tr[x].covmn = INT_MAX; } } void outputinfo(int x){ debug("Info of vertex %d:\n" ,x); debug("> val = %d ,mnval = %d ,mnvalfrom = %d\n" ,tr[x].val ,tr[x].mnval ,tr[x].mnfr); debug("> point = (%d ,%d)\n" ,tr[x].tp.x[0] ,tr[x].tp.x[1]); } int build(int L ,int R ,int d = 1){ if(L > R) return 0; int mid = (L + R) >> 1 ,x = mid; D = d ,std::nth_element(P + L ,P + mid ,P + R + 1); trans[mid] = P[mid].id; tkdt[P[mid].id] = mid; tr[x].tp = P[mid]; tr[x].covmn = INT_MAX; tr[x].val = tr[x].tp.w; lch = build(L ,mid - 1 ,!d); rch = build(mid + 1 ,R ,!d); _pushup(x); // outputinfo(x); return x; } void buildtree(int L ,int R){rt = build(L ,R);} bool inside(point A ,point B ,point C ,point D){ // check if AB is in CD for(int i : {0 ,1}){ if(!(C.x[i] <= A.x[i] && B.x[i] <= D.x[i])){ return false; } } return true; } bool outside(point A ,point B ,point C ,point D){ return B.x[0] < C.x[0] || A.x[0] > D.x[0] || B.x[1] < C.x[1] || A.x[1] > D.x[1]; } void upd(point A ,point B ,int x ,int v){ if(!x) return; pushdown(x); if(inside(tr[x].mn ,tr[x].mx ,A ,B)){return docovmn(x ,v);} if(tr[x].mxval <= v || outside(tr[x].mn ,tr[x].mx ,A ,B) || !tr[x].alive){return;} if(inside(tr[x].tp ,tr[x].tp ,A ,B) && !tr[x].del){ debug("upd (%d ,%d) by %d\n" ,tr[x].tp.x[0] ,tr[x].tp.x[1] ,v); tr[x].val = std::min(tr[x].val ,v); } upd(A ,B ,lch ,v) ,upd(A ,B ,rch ,v); return pushup(x); } void del(point A ,int x){ if(!x) return ; if(outside(A ,A ,tr[x].mn ,tr[x].mx) || !tr[x].alive) return; pushdown(x); if(inside(tr[x].tp ,tr[x].tp ,A ,A)){ tr[x].del = true; pushup(x); return ; } del(A ,lch) ,del(A ,rch); pushup(x); } std::pair<int ,int> getmn(){ return std::make_pair(tr[rt].mnval ,tr[rt].mnfr); } void pushall(int x){ if(!x) return; pushdown(x); pushall(lch) ,pushall(rch); pushup(x); outputinfo(x); } }kdt; struct Edge{ int w; point A ,B; Edge(){w = 0 ,A = B = point();} Edge(int dist ,point a ,point b){ w = dist; A = a ,B = b; } }; std::vector<Edge> e[MX]; int dis[MX]; int main(){ __FILE([NOI2019]彈跳); int n = read() ,m = read() ,w = read() ,h = read(); for(int i = 1 ,x ,y ; i <= n ; ++i){ x = read() ,y = read(); P[i] = point(x ,y ,(INT_MAX) >> 1 ,i); } kdt.buildtree(1 ,n); for(int i = 1 ,p ,t ,L1 ,R1 ,L2 ,R2 ; i <= m ; ++i){ p = read() ,t = read(); L1 = read() ,L2 = read() ,R1 = read() ,R2 = read(); e[p].push_back(Edge(t ,point(L1 ,R1) ,point(L2 ,R2))); } memset(dis ,0x3f ,sizeof dis); kdt.upd(P[tkdt[1]] ,P[tkdt[1]] ,kdt.rt ,0); for(int i = 1 ; i <= n ; ++i){ // kdt.pushall(kdt.rt); std::pair<int ,int> cur = kdt.getmn(); int x = trans[cur.second] ,dist = cur.first; debug("%d updated\n" ,x); kdt.del(P[cur.second] ,kdt.rt); dis[x] = dist; for(auto j : e[x]){ debug("$ TRY to update from %d...\n" ,x); kdt.upd(j.A ,j.B ,kdt.rt ,dist + j.w); } } for(int i = 2 ; i <= n ; ++i){ printf("%d\n" ,dis[i]); } return 0; }