HDU 3887 Counting Offspring 樹狀數組 + 棧類比dfs

來源：http://acm.hdu.edu.cn/showproblem.php?pid=3887

題意：給一棵樹，樹的根結點給出，邊也給出，現在問每個結點下面有多少個結點的編號比該結點的編號小。

思路：這道題就是POJ 3321 和 HDU 4417的結合。首先用dfs遍曆樹，對每個結點對應一個區間，然後就是求一個區間內比一個數小的數有多少個，和HDU  4417 一樣。不同的是這道題目dfs會爆棧，因此需要和棧類比dfs,或者調棧的大小。

棧類比dfs代碼：

#include <iostream>#include <cstdio>#include <string.h>#include <algorithm>#include <stdio.h>#include <stack>#include <vector>using namespace std;const int N = 100010;struct ask{int lp,rp,value;}aa[N];struct dit{int id,num;}dd[N];int n,root,cnt[N],timeorder = 0;bool vis[N];vector<int> vv[N];//void dfs(int x){//aa[x].lp = timeorder;//aa[x].value = x;//vis[x] = true;//for(int i = 0; i < vv[x].size(); ++i){//    int y = vv[x][i];//if(!vis[y])//   dfs(y);//vis[y] = true;//}//aa[x].rp = timeorder++;//}void dfs(int x){stack<int> ss;ss.push(x);while(!ss.empty()){   int tt = ss.top();   if(!vis[tt]){     vis[tt] = true; aa[tt].lp = timeorder; aa[tt].value = tt;   }   bool flag = false;   for(int i = 0; i < vv[tt].size(); ++i){   int y = vv[tt][i];   if(!vis[y]){   ss.push(y);   flag = true;   break;   }   }   if(flag) continue;   if(vis[tt]){     aa[tt].rp = timeorder++; ss.pop();   }}}bool cmp1(ask a,ask b){return a.rp < b.rp;}bool cmp2(ask a,ask b){return a.value < b.value;}int inline lowbit(int x){return x & (-x);}int inline sum(int x){int s = 0;while(x > 0){   s += cnt[x];   x -= lowbit(x);}return s;}void inline update(int x){while(x < N){//printf("ss\n");   cnt[x]++;   x += lowbit(x);}}int main(){//freopen("1.txt","r",stdin);while(scanf("%d%d",&n,&root) && (n + root)){   memset(cnt,0,sizeof(cnt));   memset(vv,0,sizeof(vv));   memset(vis,0,sizeof(vis));    for(int i = 0; i < N; ++i){      dd[i].id = dd[i].num = -1;  aa[i].lp = aa[i].rp = aa[i].value = -1;   }   int x,y;   for(int i = 1; i < n; ++i){      scanf("%d%d",&x,&y);  vv[x].push_back(y);  vv[y].push_back(x);   }   timeorder = 1;   dfs(root);   sort(aa+1,aa+n+1,cmp1);   for(int i = 1; i <= n; ++i){   int x = aa[i].value;      dd[x].id = i;  dd[i].num = aa[i].value;   }   sort(aa+1,aa+n+1,cmp2);   int ans[N] = {0};   for(int i = 1; i <= n; ++i){      ans[i] = sum(aa[i].rp) - sum(aa[i].lp - 1);  update(dd[i].id);   }   for(int i = 1; i < n; ++i)   printf("%d ",ans[i]);   printf("%d\n",ans[n]);}return 0;}

調棧大小的代碼：

#pragma comment(linker,"/STACK:100000000,100000000")#include <iostream>#include <cstdio>#include <string.h>#include <algorithm>#include <stdio.h>#include <stack>#include <vector>using namespace std;const int N = 100010;struct ask{int lp,rp,value;}aa[N];struct dit{int id,num;}dd[N];int n,root,cnt[N],timeorder = 0;bool vis[N];vector<int> vv[N];void dfs(int x){aa[x].lp = timeorder;aa[x].value = x;vis[x] = true;for(int i = 0; i < vv[x].size(); ++i){    int y = vv[x][i];if(!vis[y])   dfs(y);vis[y] = true;}aa[x].rp = timeorder++;}bool cmp1(ask a,ask b){return a.rp < b.rp;}bool cmp2(ask a,ask b){return a.value < b.value;}int inline lowbit(int x){return x & (-x);}int inline sum(int x){int s = 0;while(x > 0){   s += cnt[x];   x -= lowbit(x);}return s;}void inline update(int x){while(x < N){//printf("ss\n");   cnt[x]++;   x += lowbit(x);}}int main(){//freopen("1.txt","r",stdin);while(scanf("%d%d",&n,&root) && (n + root)){   memset(cnt,0,sizeof(cnt));   memset(vv,0,sizeof(vv));   memset(vis,0,sizeof(vis));    for(int i = 0; i < N; ++i){      dd[i].id = dd[i].num = -1;  aa[i].lp = aa[i].rp = aa[i].value = -1;   }   int x,y;   for(int i = 1; i < n; ++i){      scanf("%d%d",&x,&y);  vv[x].push_back(y);  vv[y].push_back(x);   }   timeorder = 1;   dfs(root);   sort(aa+1,aa+n+1,cmp1);   for(int i = 1; i <= n; ++i){   int x = aa[i].value;      dd[x].id = i;  dd[i].num = aa[i].value;   }   sort(aa+1,aa+n+1,cmp2);   int ans[N] = {0};   for(int i = 1; i <= n; ++i){      ans[i] = sum(aa[i].rp) - sum(aa[i].lp - 1);  update(dd[i].id);   }   for(int i = 1; i < n; ++i)   printf("%d ",ans[i]);   printf("%d\n",ans[n]);}return 0;}