标签:har const main break size empty 元组 mes jks
kruskal 重构树
对于一张无向图,我们在进行 kruskal 的过程中
每当合并两个联通块时
新建虚拟节点 t
对于两个联通块的根节点 fau,fav 连无向边
(fau, t),(fav, t) 其中点 t 的点权为两个联通块当前连边的边权
对于这道题
首先 dijkstra 处理所有点到1号点的最短路
然后按照边的海拔进行降序排序
这样做出重构树之后
显然对于点 u,它的所有子树中的相关的边的海拔(这里已经转化为了虚拟节点的点权)都要大于该点的海拔
这样的话
对于询问二元组 x, h
倍增将 x 调到海拔最低且高于 h 的点处
此时 x 的子树中dis[]的最小值即为此次询问的结果
注意:在进行重构树时
虚拟节点的dis[]每次可以取 min(dis[fau], dis[fav])
这样就相当于dis[t]表示 t 的子树中dis[]的最小值
省去了一遍 dfs
#include <iostream> #include <cstdio> #include <algorithm> #include <queue> #include <cstring> using namespace std; const int N = 4e5 + 10, oo = 1e9 + 7; struct Node { int u, v, len, high, nxt; } E[N], G[N << 2], Edge[N << 1]; struct Node_ { int u, dis_; inline bool operator < (const Node_ a) const {return dis_ > a.dis_;} }; int head_1[N], head_2[N << 1], now; int dis[N << 1]; bool vis[N]; int fa[N << 1]; int n, m; int High[N << 1]; int f[N << 1][30]; int deep[N << 1]; #define gc getchar() inline int read() { int x = 0; char c = gc; while(c < ‘0‘ || c > ‘9‘) c = gc; while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = gc; return x; } int Get(int x) {return fa[x] == x ? x : fa[x] = Get(fa[x]);} inline bool Cmp(Node a, Node b) {return a.high > b.high;} void Dfs(int u, int fa) {for(int i = head_2[u]; ~ i; i = G[i].nxt) if(G[i].v != fa) f[G[i].v][0] = u, Dfs(G[i].v, u);} inline int Jump(int X, int H) {for(int i = 20; i >= 0; i --) if(f[X][i] && High[f[X][i]] > H) X = f[X][i];return X;} inline void Add_Edge(int u, int v, int Len) {Edge[++ now].v = v; Edge[now].len = Len; Edge[now].nxt = head_1[u]; head_1[u] = now;} inline void Add_G(int u, int v) {G[++ now].v = v; G[now].nxt = head_2[u]; head_2[u] = now;} inline void Pre() {for(int i = 1; i <= 20; i ++) for(int j = 1; j <= (n * 2 - 1); j ++) f[j][i] = f[f[j][i - 1]][i - 1];} inline void Dijkstra() { for(int i = 1; i <= n; i ++) dis[i] = oo; for(int i = 1; i <= n; i ++) vis[i] = 0; priority_queue <Node_> Q; Q.push((Node_) {1, 0}); dis[1] = 0; while(!Q.empty()) { Node_ topp = Q.top(); Q.pop(); if(vis[topp.u]) continue; vis[topp.u] = 1; for(int i = head_1[topp.u]; ~ i; i = Edge[i].nxt) if(dis[Edge[i].v] > dis[topp.u] + Edge[i].len) { dis[Edge[i].v] = dis[topp.u] + Edge[i].len; Q.push((Node_) {Edge[i].v, dis[Edge[i].v]}); } } } inline void Kruskal() { sort(E + 1, E + m + 1, Cmp); for(int i = 1; i <= (n << 1); i ++) fa[i] = i; for(int i = 1; i <= (n << 1); i ++) head_2[i] = -1; int cnt = n; now = 0; for(int i = 1; i <= m; i ++) { if(cnt == n * 2 - 1) break; int u = E[i].u, v = E[i].v, fau = Get(u), fav = Get(v); if(fau != fav) { fa[fau] = fa[fav] = ++ cnt; High[cnt] = E[i].high; dis[cnt] = min(dis[fau], dis[fav]); Add_G(fau, cnt), Add_G(cnt, fau), Add_G(fav, cnt), Add_G(cnt, fav); } } } int main() { for(int T = read(); T; T --) { memset(f, 0, sizeof f); n = read(), m = read(); now = 0; for(int i = 1; i <= n; i ++) head_1[i] = -1; for(int i = 1; i <= m; i ++) { int u = read(), v = read(), Len = read(), high = read(); Add_Edge(u, v, Len), Add_Edge(v, u, Len); E[i] = (Node) {u, v, Len, high}; } Dijkstra(), Kruskal(), Dfs(2 * n - 1, 0), Pre(); int Q = read(), K = read(), S = read(), Lastans = 0; for(; Q; Q --) { int X = (read() + K * Lastans - 1) % n + 1, H = (read() + K * Lastans) % (S + 1); Lastans = dis[Jump(X, H)]; cout << Lastans << "\n"; } } return 0; }
标签:har const main break size empty 元组 mes jks
原文地址:https://www.cnblogs.com/shandongs1/p/9497615.html