HDU 5692 Segment Tree

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Test instructions: Chinese title

Idea: This is the first time to write such a line segment tree, personal feeling is a routine has not done before, but thought what ingenious, will a tree to transform into a line tree really powerful, first will each node below all the number of DFS serial number, and will their left and right number down, that is, my Code L and R, The Val record is the root node 0 to the current location of the cost, after the achievement is very good operation, the update will find the current node's son and their own representative of the interval, and then update the interval, note that the topic Update a node value into Y, why we update a section, because the node of our segment tree is saved to the root node 0 , then as a Father node, its value changed, will naturally affect its son to the root of the cost of the node, and change the same, so with the interval update to complete, the query is the same, must go to the X node, then to the X node is not there will be a greater cost, and this cost of the point must go through the X node, So we directly query the X node and all of its son's maximum value can PS: tree-shaped structure and line segment tree perfect combination, to practice well

#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include <vector> #include <stdio.h> #include < string.h> #include <stdlib.h> #include <iostream> #include <algorithm>using namespace Std;typedef Long Long ll;const ll Inf=0x3f3f3f3f3f3f3f3f;const int Maxn=100010;vector<int>g[maxn];ll MAX1[MAXN*4],LAZY[MAXN    *4],val[maxn];int l[maxn],r[maxn],cost[maxn];int idx;void dfs (int u,int f,ll vv) {l[u]=++idx;        for (unsigned int i=0;i<g[u].size (); i++) {int v=g[u][i];        if (v==f) continue;    DFS (V,u, (ll) COST[V]+VV);    } R[u]=idx; VAL[L[U]]=VV;} void pushup (int node) {Max1[node]=max (max1[node<<1],max1[node<<1|1]);}        void pushdown (int node) {if (Lazy[node]) {lazy[node<<1]+=lazy[node];        lazy[node<<1|1]+=lazy[node];        max1[node<<1]+=lazy[node];        max1[node<<1|1]+=lazy[node];    lazy[node]=0; }}void buildtree (int le,int ri,int node) {if (Le==ri) {Max1[node]=vAl[le];    return;    } int t= (LE+RI) >>1;    Buildtree (le,t,node<<1);    Buildtree (t+1,ri,node<<1|1); Pushup (node);}        void update (int l,int r,int x,int le,int ri,int node) {if (l<=le&&ri<=r) {lazy[node]+= (ll) x;        max1[node]+= (ll) x;    return;    } pushdown (node);    int t= (LE+RI) >>1;    if (l<=t) update (L,R,X,LE,T,NODE&LT;&LT;1);    if (r>t) update (L,R,X,T+1,RI,NODE&LT;&LT;1|1); Pushup (node);}    ll query (int l,int r,int le,int ri,int node) {if (l<=le&&ri<=r) return Max1[node];    Pushdown (node);    ll Ans=-inf;    int t= (LE+RI) >>1;    if (l<=t) Ans=max (Ans,query (l,r,le,t,node<<1));    if (r>t) Ans=max (Ans,query (l,r,t+1,ri,node<<1|1)); return ans;}    int main () {int t,cas=1,n,m,a,b,c;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n,&m);        for (int i=0;i<maxn;i++) g[i].clear (); for (int i=0;i<n-1;i++) {scanf ("%d%d", &a,&b);            a++;b++;            G[a].push_back (b);        G[b].push_back (a);        } for (int i=1;i<=n;i++) scanf ("%d", &cost[i]);        Idx=0;dfs (1,0, (LL) cost[1]);        Buildtree (1,n,1);        memset (lazy,0,sizeof (lazy));        printf ("Case #%d:\n", cas++);            while (m--) {scanf ("%d", &a);                if (a==0) {scanf ("%d%d", &b,&c); b++;                Update (l[b],r[b],c-cost[b],1,n,1);            Cost[b]=c;                }else{scanf ("%d", &b); b++;                ll Answe=query (l[b],r[b],1,n,1);            printf ("%i64d\n", answe); }}} return 0;}

HDU 5692 Segment Tree