NEFU 1160 线段树

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题意：弱校的OJ的弱题，给定一棵树，和关系，然后在节点上更新时，它的所有儿子也要更新，有个判断条件，查询是所有节点的值的和

思路：与HDU 5692一样的思路，不多说了

#pragma comment(linker, "/STACK:102400000, 102400000")
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int maxn=100010;
vectorG[maxn];
ll num[maxn*4],lazy[maxn*4],min1[maxn*4];
int L[maxn],R[maxn];
int idx;
void dfs(int u,int f){
    L[u]=++idx;
    for(unsigned int i=0;i>1));
        num[node<<1|1]+=lazy[node]*(ll)(m>>1);
        min1[node<<1]+=lazy[node];
        min1[node<<1|1]+=lazy[node];
        lazy[node]=0;
    }
}
void buildtree(int le,int ri,int node){
    if(le==ri){
        min1[node]=num[node]=1;
        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,ll x,int le,int ri,int node){
    if(l<=le&&ri<=r){
        lazy[node]+=x;
        num[node]+=x*(ll)(ri-le+1);
        min1[node]+=x;
        return ;
    }
    pushdown(node,ri-le+1);
    int t=(le+ri)>>1;
    if(l<=t) update(l,r,x,le,t,node<<1);
    if(r>t) update(l,r,x,t+1,ri,node<<1|1);
    pushup(node);
}
ll query_sum(int l,int r,int le,int ri,int node){
    if(l<=le&&ri<=r) return num[node];
    pushdown(node,ri-le+1);
    ll ans=0;
    int t=(le+ri)>>1;
    if(l<=t) ans+=query_sum(l,r,le,t,node<<1);
    if(r>t) ans+=query_sum(l,r,t+1,ri,node<<1|1);
    return ans;
}
ll query_min(int l,int r,int le,int ri,int node){
    if(l<=le&&ri<=r) return min1[node];
    pushdown(node,ri-le+1);
    ll ans=999999999999ll;
    int t=(le+ri)>>1;
    if(l<=t) ans=min(ans,query_min(l,r,le,t,node<<1));
    if(r>t) ans=min(ans,query_min(l,r,t+1,ri,node<<1|1));
    return ans;
}
int main(){
    int T,cas=1,n,m,a,b,c;
    char ch[10];
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=0;i