3351: [ioi2009]Regions

3351: [ioi2009]Regions

Time Limit: 120 Sec  Memory Limit: 128 MB
Submit: 205  Solved: 59
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Description

 

N个节点的树，有R种属性，每个点属于一种属性。有Q次询问，每次询问r1,r2，回答有多少对(e1,e2)满足e1属性是r1，e2属性是r2，e1是e2的祖先。

数据规模
N≤200000,R≤25000，Q≤200000

30%数据R≤500

55%数据同种属性节点个数≤500

Input

Output

Sample Input

6 3 4
1
1 2
1 3
2 3
2 3
5 1
1 2
1 3
2 3
3 1

Sample Output

1
3
2
1

HINT

Source

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参考于《根号算法——不只是分块》 王悦同

设属性r1的点有A个属性r2的点有B个

若A很小，则设计一个AlogB的算法，若B很小，则设计一个BlogA的算法

若都很大，这样的询问不会很多，就直接O(N)暴力

针对不同类型询问设计不同的专杀算法，运用根号卡时间

苟蒻编写能力还有待提高！注意数组下标。。。

#include<iostream>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#include<cmath>#include<vector>#include<queue>using namespace std;const int maxn = 2E5 + 20;struct Q{int r1,r2,num;bool operator < (const Q &b) const {if (r1 < b.r1) return 1;if (r1 > b.r1) return 0;return r2 < b.r2;}bool operator == (const Q &b) const {return (r1 == b.r1 && r2 == b.r2);}}q[maxn];struct P{int pos,add,sum;bool operator < (const P &b) const {return pos < b.pos;}};vector <int> v[maxn],v2[maxn];vector <P> v3[maxn];int mark[maxn],dfs[maxn],n,m,ans[maxn],dfs_clock = 0,flag;int head[maxn],tail[maxn];int getint(){int ret = 0;char ch = getchar();while (ch < '0' || ch > '9') ch = getchar();while ('0' <= ch && ch <= '9') {ret = ret*10 + ch - '0';ch = getchar();}return ret;}void DFS(int k){dfs[k] = head[k] = ++dfs_clock;v2[mark[k]].push_back(k);for (int i = 0; i < v[k].size(); i++) DFS(v[k][i]);tail[k] = dfs_clock;if (head[k] != tail[k]) {v3[mark[k]].push_back((P){head[k],1,1});v3[mark[k]].push_back((P){tail[k]+1,-1,-1});}}int DBSC1(int x,int l,int r,int pos){if (r - l <= 1) {if (dfs[v2[x][l]] >= pos) return l;else return r;}int mid = (l+r) >> 1;if (dfs[v2[x][mid]] >= pos) return DBSC1(x,l,mid,pos);else return DBSC1(x,mid,r,pos);}int DBSC2(int x,int l,int r,int pos){if (r - l <= 1) {if (dfs[v2[x][r]] <= pos) return r;else return l;} int mid = (l+r) >> 1;if (dfs[v2[x][mid]] <= pos) return DBSC2(x,mid,r,pos);else return DBSC2(x,l,mid,pos);}int DBSC(int x,int l,int r,int pos){if (r - l <= 1) {if (v3[x][r].pos <= pos) return r;else return l;} int mid = (l+r) >> 1;if (v3[x][mid].pos <= pos) return DBSC(x,mid,r,pos);else return DBSC(x,l,mid,pos);}void solve1(int x){int ANS = 0;int r1 = q[x].r1;int r2 = q[x].r2;for (int i = 0; i < v2[r1].size(); i++) {int now = v2[r1][i];if (head[now] == tail[now]) continue;if (tail[now] < dfs[v2[r2][0]]) continue;if (head[now] > dfs[v2[r2][v2[r2].size()-1]]) continue;int L = DBSC1(r2,0,v2[r2].size()-1,head[now]);int R = DBSC2(r2,0,v2[r2].size()-1,tail[now]);ANS += R-L+1;}ans[q[x].num] = ANS;}void solve2(int x){int ANS = 0;int r1 = q[x].r1;int r2 = q[x].r2;for (int i = 0; i < v2[r2].size(); i++) {int now = v2[r2][i];if (dfs[now] < v3[r1][0].pos) continue;if (dfs[now] > v3[r1][v3[r1].size()-1].pos) continue;/*for (int i = 0; i < v3[r1].size(); i++) {int pos = v3[r1][i].pos;int add = v3[r1][i].add;int sum = v3[r1][i].sum;int bbbb = 1;}*/int POS = DBSC(r1,0,v3[r1].size()-1,dfs[now]);ANS += v3[r1][POS].sum;}ans[q[x].num] = ANS;}void solve3(int x){int ANS,SUM,L,R;int r1 = q[x].r1;int r2 = q[x].r2;ANS = SUM = L = R = 0;while (L < v3[r1].size() && R < v2[r2].size()) {if (v3[r1][L].pos <= dfs[v2[r2][R]]) {SUM += v3[r1][L].add;L++; }else {ANS += SUM;R++;}}//while (R < v2[r2].size()) ANS += SUM,R++;ans[q[x].num] = ANS;}int main(){#ifdef YZY   freopen("yzy.txt","r",stdin);#endifint tt;cin >> n >> tt >> m >> mark[1];flag = sqrt(n);for (int i = 2; i <= n; i++) {int x,y;x = getint(); y = getint();v[x].push_back(i); mark[i] = y;}DFS(1);/*for (int i = 1; i <= tt; i++) {for (int j = 0; j < v2[i].size(); j++) {int kk = v2[i][j];int b = 1;}for (int j = 0; j < v3[i].size(); j++) {int pos = v3[i][j].pos;int add = v3[i][j].add;int sum = v3[i][j].sum;int b = 1;}}*/for (int i = 1; i <= tt; i++) {sort(v3[i].begin(),v3[i].end());for (int j = 1; j < v3[i].size(); j++) v3[i][j].sum += v3[i][j-1].sum;}for (int i = 1; i <= m; i++) {int x,y;x = getint(); y = getint();q[i] = (Q){x,y,i};}sort(q+1,q+m+1);for (int i = 1; i <= m; i++) {if (q[i] == q[i-1]) {ans[q[i].num] = ans[q[i-1].num];continue;}int S1 = v2[q[i].r1].size();int S2 = v2[q[i].r2].size();if (S1 <= flag && S2 <= flag) {solve3(i);continue;}if (S1 <= flag) {solve1(i);continue;}if (S2 <= flag) {solve2(i);continue;}solve3(i);}for (int i = 1; i <= m; i++) printf("%d\n",ans[i]);return 0;}