SPOJ 375 树链剖分

新模板，配合新线段树版本了：

#include <bits/stdc++.h>#include <ext/pb_ds/priority_queue.hpp>#include <tr1/unordered_map>using std::tr1::unordered_map;//using std::setiosflags;//using std::setprecision;using std::sort;using std::max;using std::min;using std::cout;using std::stack;using std::cin;using std::endl;using std::swap;using std::pair;using std::vector;using std::set;using std::map;using std::make_pair;using std::multiset;using std::unique;using std::queue;using std::greater;using std::string;using std::priority_queue;using std::lower_bound;//返回第一个不小于using std::upper_bound;//返回第一个大于using std::max_element;using std::min_element;using __gnu_pbds::pairing_heap_tag;#define Hash unordered_map#define clr(x) memset(x,0,sizeof(x))#define logz(x) (31-__builtin_clz((int)x))#define logzl(x) (63-__builtin_clzll((LL)x))#define cf std::ios::sync_with_stdio(0);std::cin.tie(0)typedef unsigned long long uLL;typedef long long LL;const double PI = acos(-1);#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>using namespace std;const int maxn = 100100;const int inf = 0x7fffffff;//----建边部分struct Vector_edge//存vector用的东西{int will, w;//v为连接的点， w为边权Vector_edge():will(0),w(0){}Vector_edge(int a, int b):will(a),w(b){}};vector<Vector_edge>g[maxn];struct edge{int to, next;}e[maxn];inline void add_edge(int a, int b, int c){g[a].push_back(Vector_edge(b,c));}//----树链部分=====int dfs_clock;//DFS时间序列struct node{int size, dep, son, top, fa, ti;node():size(0),dep(0),son(0),top(0),fa(0),ti(0){};//size:v为跟节点数量//dep:v的深度 根为1//top:v所在链的顶端//son:vv在同一重链的儿子节点，重儿子//ti:v与父亲的连边，在线段树中的位置}t[maxn];void dfs_find(int u, int pa, int depth)//第一次DFS，得到size,dep,fa以及son的数据{t[u].son = 0;//重儿子，为0表示没有重儿子t[u].size = 1;//节点数量t[u].dep = depth;t[u].fa = pa;for (int i = 0; i != g[u].size(); ++ i){int will = g[u][i].will;if (will == pa)continue;dfs_find(will, u, depth + 1);t[u].size += t[will].size;if (t[will].size > t[t[u].son].size)t[u].son = will;}}void dfs_time(int u, int pa)// 得到时间戳等数据,u为当前节点，pa为父链顶端节点{t[u].ti = ++ dfs_clock; //u这个节点的时间戳是dfs_clock,dfs_clock下标是从1开始的，没有0！t[u].top = pa;//top是u所在链的顶端if (t[u].son != 0)dfs_time(t[u].son, t[u].top);//如果节点有重儿子，那么依旧是以pa为链顶端的一条链for (int i = 0; i != g[u].size(); ++ i){int will = g[u][i].will;if (will == t[u].son || will == t[u].fa)continue;//重儿子或者父节点，则跳过dfs_time(will, will);//新的一条链}}//----线段树部分--int val[maxn<<2], mx[maxn<<2];#define lson o*2, L, M  #define rson o*2+1, M + 1,R  int segment_tree_query_l, segment_tree_query_r, segment_tree_query_ans;//int segment_tree_update_l, segment_tree_update_r, segment_tree_update_val;//这种线段树写法，需要把询问的东西，放在全局变量//qans就是query询问的结果，ql,qr是询问[L,R]区间的答案//ql,qr,qans对于update的时候，本着不浪费的原则，复用一下，就不用upl,upr,和upval了#define qans segment_tree_query_ans#define ql segment_tree_query_l#define qr segment_tree_query_r//upl ,upr是update的[L,R]区间，改为upval#define upl segment_tree_update_l#define upr segment_tree_update_r#define upval segment_tree_update_valint n;void update(int o, int L, int R)// L,R区间统一改为qans{if (ql <= L && R <= qr){val[o] = mx[o] = qans; return ;}int M = L + (R - L) / 2;if (ql <= M) update(lson);if (qr > M)update(rson);mx[o] = max(mx[o*2], mx[o*2+1]);}void query(int o, int L, int R){if (ql <= L && R <= qr){qans= max(qans, mx[o]);return ;}int M = L + (R- L)/2;if (ql <= M)query(lson);if (qr > M)query(rson);}int lca(int x, int y)//找出x到y的之间所有的区间{int ans = -inf;//cout<<t[x].top <<" " << t[y].top<<endl;while (t[x].top != t[y].top){if (t[t[x].top].dep < t[t[y].top].dep)swap(x,y);//x深，y浅//线段树查询beginqans= -inf;//切记要初始化qansql=t[t[x].top].ti;qr=t[x].ti;query(1,1,n);ans = max(ans, qans);//线段树查询endx = t[t[x].top].fa;}if (t[x].dep > t[y].dep)swap(x, y);//x是上面一些的节点if (x != y){//线段树查询beginqans= -inf;ql=t[x].ti + 1;//一定要+1，因为x,y在一条链上，所以标号连续。qr=t[y].ti;query(1,1,n);ans = max(ans, qans);//线段树查询end}return ans;}int da[maxn][3];//起点, 终点, 权重void clear(){dfs_clock = 0;for (int i = 0; i != maxn; ++ i)g[i].clear();memset(mx, 0, sizeof(mx));memset(val, 0, sizeof(val));}int main(){int T; scanf("%d", &T);while (T--){clear();scanf("%d", &n);for (int i = 1; i < n; ++ i){scanf("%d%d%d", &da[i][0], &da[i][1], &da[i][2]);add_edge(da[i][0], da[i][1], da[i][2]);add_edge(da[i][1], da[i][0], da[i][2]);}dfs_find(1, 1, 1);dfs_time(1, 1);for (int i = 1; i < n; i ++ )//人工更新所有的边{if (t[da[i][0]].dep > t[da[i][1]].dep)//相邻的边，一定是父子关系swap(da[i][0] , da[i][1]);//da[i][1]是深度深的点, 然后da[i][1]这个点的DFS序号，就是他的前驱边的序号//线段树update  beginql = t[da[i][1]].ti;qr = t[da[i][1]].ti;qans = da[i][2];update(1,1,n);//线段树update  end}char he[100];while(1){scanf("%s",he);if(he[0]=='D')break;if(he[0]=='Q'){int a,b;scanf("%d%d",&a,&b);printf("%d\n",lca(a,b));}else{int a,b;scanf("%d%d",&a,&b);//线段树update  beginql = t[da[a][1]].ti;qr = t[da[a][1]].ti;qans = b;update(1,1,n);//线段树update  end}}printf("\n");}return 0;}/*   1   5   1 2 3   1 3 1   3 4 3   1 5 3   QUERY 1 4   DONE*/

模板

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>using namespace std;#define lson (pos<<1)#define rson (pos<<1|1)const int maxn = 100100;const int inf = 0x7fffffff;//----建边部分struct Vector_edge//存vector用的东西{int v, w;Vector_edge():v(0),w(0){}Vector_edge(int a, int b):v(a),w(b){}};vector<Vector_edge>g[maxn];struct edge{int to, next;}e[maxn];inline void add_edge(int a, int b, int c){g[a].push_back(Vector_edge(b,c));}//----树链部分=====int dfs_clock, cnt;struct node{int size, dep, son, top, fa, ti;node():size(0),dep(0),son(0),top(0),fa(0),ti(0){};//size:v为跟节点数量//dep:v的深度 根为1//top:v所在链的顶端//son:vv在同一重链的儿子节点，重儿子//ti:v与父亲的连边，在线段树中的位置}t[maxn];void dfs_find(int u, int pa, int depth){//cout<<u<<" "<<pa<<" "<<depth<<endl;t[u].son = 0;t[u].size = 1;t[u].dep = depth;t[u].fa = pa;for (int i = 0; i != g[u].size(); ++ i){int v = g[u][i].v;if (v == pa)continue;dfs_find(v, u, depth + 1);t[u].size += t[v].size;//if (t[u].son == -1)t[u].son = v;if (t[v].size > t[t[u].son].size)t[u].son = v;}}void dfs_time(int u, int pa){t[u].ti = ++ dfs_clock; t[u].top = pa;if (t[u].son != 0)dfs_time(t[u].son, t[u].top);for (int i = 0; i != g[u].size(); ++ i){int v = g[u][i].v;if (v == t[u].son || v == t[u].fa)continue;dfs_time(v, v);}}//----线段树部分--struct xianduan_node{int l,r,val,mx;int mid(){return (l+r)>>1;}}tree[maxn<<2];void build(int pos, int l, int r){tree[pos].l = l;tree[pos].r = r;tree[pos].mx = tree[pos].val = - inf;if (l == r)return;int mid = tree[pos].mid();build(lson,l,mid);build(rson, mid + 1, r);}void update(int pos, int id, int w)// id改为w{if (tree[pos].l == tree[pos].r){tree[pos].mx = tree[pos].val = w;return ;}int mid = tree[pos].mid();if (mid >= id)update(lson, id, w);else update(rson, id, w);tree[pos].mx = max(tree[lson].mx, tree[rson].mx);}int query(int pos, int L, int R){//cout<<L<<" "<<R<<" "<<pos<<" "<<tree[pos].l<<" "<<tree[pos].r<<" "<<tree[pos].mx<<endl;if (L <= tree[pos].l && tree[pos].r <= R)return tree[pos].mx;int mid = tree[pos].mid();int ans = - inf;if (mid >= L)ans = max(ans, query(lson, L, R));if (mid < R)ans = max(ans, query(rson, L, R));return ans;}int lca(int x, int y){int ans = -inf;while (t[x].top != t[y].top){if (t[t[x].top].dep < t[t[y].top].dep)swap(x,y);ans = max(ans, query(1, t[t[x].top].ti, t[x].ti));x = t[t[x].top].fa;}if (t[x].dep > t[y].dep)swap(x, y);//cout<<x<<" "<<y<<endl;//cout<<ans<<endl;//cout<<t[x].ti<<" "<<t[y].ti<<endl;//cout<<endl<<endl;if (x != y)ans = max(ans , query(1, t[x].ti + 1, t[y].ti));////这里的t[x].ti指的是x与其父亲的边，所以一定注意+1return ans;}int n;int da[maxn][3];void clear(){cnt = dfs_clock = 0;for (int i = 0; i != maxn; ++ i)g[i].clear();}int main(){int T; scanf("%d", &T);while (T--){clear();scanf("%d", &n);for (int i = 1; i < n; ++ i){scanf("%d%d%d", &da[i][0], &da[i][1], &da[i][2]);add_edge(da[i][0], da[i][1], da[i][2]);add_edge(da[i][1], da[i][0], da[i][2]);}dfs_find(1, 1, 1);dfs_time(1, 1);build(1, 2, n);//因为1没有前驱边，1是树根for (int i = 1; i < n; i ++ ){if (t[da[i][0]].dep > t[da[i][1]].dep)swap(da[i][0] , da[i][1]);update(1, t[da[i][1]].ti, da[i][2]);}        char he[100];        while(1)        {            scanf("%s",he);            if(he[0]=='D')break;            if(he[0]=='Q')            {                int a,b;scanf("%d%d",&a,&b);                printf("%d\n",lca(a,b));            }            else            {                int a,b;scanf("%d%d",&a,&b);                update(1,t[da[a][1]].ti,b);            }        }        printf("\n");}return 0;}/*151 2 31 3 13 4 31 5 3QUERY 1 4DONE*/