BZOJ 2212线段树的合并

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借鉴（抄）了一下题解……
线段树合并的裸题吧…

//By SiriusRen#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define N 4000050typedef long long LL; int n,cnt,tree[N],son[N][2],root,Root[N],all,tr[N],s[N][2];LL ans,ans1,ans2;void build(int &x){    x=++cnt;    scanf("%d",&tree[x]);    if(tree[x])return;    build(son[x][0]),build(son[x][1]);}void dfs(int x){    if(!x)return;    printf("x=%d tree[x]=%d\n",x,tree[x]);    dfs(son[x][0]),dfs(son[x][1]);}void push_up(int x){tr[x]=tr[s[x][0]]+tr[s[x][1]];}void insert(int &x,int l,int r,int wei){    if(!x)x=++all;    if(l==r){tr[x]=1;return;}    int mid=(l+r)>>1;    if(wei<=mid)insert(s[x][0],l,mid,wei);    else insert(s[x][1],mid+1,r,wei);    push_up(x);}int merge(int x,int y){    if(!x)return y;if(!y)return x;    ans1+=1LL*tr[s[x][1]]*tr[s[y][0]];    ans2+=1LL*tr[s[x][0]]*tr[s[y][1]];    s[x][0]=merge(s[x][0],s[y][0]);    s[x][1]=merge(s[x][1],s[y][1]);    push_up(x);return x;}void solve(int x){    if(tree[x])return;    solve(son[x][0]),solve(son[x][1]);    ans1=ans2=0;    Root[x]=merge(Root[son[x][0]],Root[son[x][1]]);    ans+=min(ans1,ans2);}int main(){    scanf("%d",&n);    build(root);    for(int i=1;i<=cnt;i++)if(tree[i])insert(Root[i],1,n,tree[i]);    solve(root);    printf("%lld\n",ans);}

这里写图片描述

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