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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#869987#9676. Ancestorsminiworld0 0ms0kbC++145.5kb2025-01-25 14:17:072025-01-25 14:17:18

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你现在查看的是最新测评结果

  • [2025-01-25 14:17:18]
  • 评测
  • 测评结果:0
  • 用时:0ms
  • 内存:0kb
  • [2025-01-25 14:17:07]
  • 提交

answer

#include<bits/stdc++.h>
#define N 1000005
#define M 20000007
using namespace std;
#define getchar() (p1 == p2 && (p2 = (p1 = buf1) + fread(buf1, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf1[1 << 23], *p1 = buf1, *p2 = buf1, ubuf[1 << 23], *u = ubuf;
namespace IO
{
	template<typename T>
	void read(T &_x){_x=0;int _f=1;char ch=getchar();while(!isdigit(ch)) _f=(ch=='-'?-1:_f),ch=getchar();while(isdigit(ch)) _x=_x*10+(ch^48),ch=getchar();_x*=_f;}
	template<typename T,typename... Args>
	void read(T &_x,Args&...others){Read(_x);Read(others...);}
	const int BUF=20000000;char buf[BUF],to,stk[32];int plen;
	#define pc(x) buf[plen++]=x
	#define flush(); fwrite(buf,1,plen,stdout),plen=0;
	template<typename T>inline void print(T x){if(!x){pc(48);return;}if(x<0) x=-x,pc('-');for(;x;x/=10) stk[++to]=48+x%10;while(to) pc(stk[to--]);}
}
using namespace IO;

int n, m, B, root;
vector<int>p[N];
int dep[N], ans[N], dfn[N], len, tot, n1, n2;

struct RMQ_LCA { //O(n log n) - O(1) LCA
    int z[N], r[N], lg[N], tot;
    int f[20][200010];
    void dfs(int x, int fa) {
        dep[x] = dep[fa] + 1;
        r[++tot] = x;
        dfn[x] = ++len;
        z[x] = tot;
        for(auto y : p[x]) {
            dfs(y, x);
            r[++tot] = x;
        }
    }
    inline int Min(int x, int y) {
        return dep[x] < dep[y] ? x : y;
    }
    inline int LCA(int x, int y) {
        x = z[x], y = z[y];
        if(x > y) swap(x, y);
        int h = lg[y - x + 1];
        return Min(f[h][x], f[h][y - (1 << h) + 1]);
    }
    void init() {
        dfs(root, 0);
        for(int i = 2; i <= tot; i++) lg[i] = lg[i / 2] + 1;
        for(int i = 1; i <= tot; i++) f[0][i] = r[i];
        for(int j = 1; j <= lg[tot]; j++) 
        for(int i = 1; i + (1 << j) - 1 <= tot; i++) f[j][i] = Min(f[j - 1][i], f[j - 1][i + (1 << (j - 1))]);
    }
}ljm;

int tim1, tim2;
struct DS { //分块:O(1)修改, O(sqrt(n))查询
    int v1[N], v2[N];
    int X[M], Y[M];
    int Get(int x) {
        return ceil((double)x * 1.0 / B);
    }
    void update(int x, int y) {
        X[++tim1] = x; Y[tim1] = y;
        v1[Get(x)] += y;
        v2[x] += y;
    }
    int query(int L, int R) {
        int res = 0;
        for(int i = L; i <= R; ) {
            if(i % B == 1 && i + B - 1 <= R) res += v1[Get(i)], i += B;
          else 
                res += v2[i], i++;
        }
        return res;
    }
    void Go(int x) { //撤销回到时间轴为x时
        while(tim1 > x) {
            v1[Get(X[tim1])] -= Y[tim1];
            v2[X[tim1]] -= Y[tim1];
            tim1--;
        }
    }
}ds;

struct List {
    int Pre, Nxt;
}t[N];

struct node {
    int l, r, x, id;
}z[N], h[N];
inline bool cmp1(node A, node B) {
    return A.l < B.l;
}
inline bool cmp2(node A, node B) {
    return A.r > B.r;
}

int v[N];
inline bool cmp(int x, int y) {
    return dfn[x] < dfn[y];
}
vector<int>r[N];
void Prepare() {
    for(int i = 1; i <= n; i++) v[i] = i;
    sort(v + 1, v + n + 1, cmp);
    for(int i = 1; i <= n; i++) r[dep[v[i]]].push_back(v[i]);
    for(int d = 1; d <= n; d++) {
        for(int i = 0; i < r[d].size(); i++) {
            if(i == 0) t[r[d][i]].Pre = -1;
            else t[r[d][i]].Pre = r[d][i - 1];
            if(i == r[d].size() - 1) t[r[d][i]].Nxt = -1;
            else {
                t[r[d][i]].Nxt = r[d][i + 1];
                ds.update(d - dep[ljm.LCA(r[d][i], r[d][i + 1])], 1);
            }
        }
        if(r[d].size()) ds.update(d, 1); //d深度只要有一个,贡献就要加1
    }
    n1 = tim1, n2 = tim2;
}


int X[M], Y[M], opt[M];
void Go(int x) {
    while(tim2 > x) {
        if(opt[tim2] == 1) t[X[tim2]].Pre = Y[tim2];
        else t[X[tim2]].Nxt = Y[tim2];
        tim2--;
    }
}
void Del(int x) {
    if(t[x].Pre != -1) {
        ds.update(dep[x] - dep[ljm.LCA(x, t[x].Pre)], -1);
        tim2++; X[tim2] = t[x].Pre, Y[tim2] = t[t[x].Pre].Nxt, opt[tim2] = 2;
        t[t[x].Pre].Nxt = t[x].Nxt; 
    }
    if(t[x].Nxt != -1) {
        ds.update(dep[x] - dep[ljm.LCA(x, t[x].Nxt)], -1);
        tim2++; X[tim2] = t[x].Nxt, Y[tim2] = t[t[x].Nxt].Pre, opt[tim2] = 1;
        t[t[x].Nxt].Pre = t[x].Pre; 
    }
    if(t[x].Pre != -1 && t[x].Nxt != -1) 
        ds.update(dep[x] - dep[ljm.LCA(t[x].Pre, t[x].Nxt)], 1);
    if(t[x].Pre == -1 && t[x].Nxt == -1)  //这个深度删掉没有点了
        ds.update(dep[x], -1);
}

void Sol(int L, int R) { //处理左端点在[L,R]的询问
    sort(h + 1, h + tot + 1, cmp2);
    for(int i = 1; i < L; i++) Del(i);
    int curL = L, curR = n;
    for(int i = 1; i <= tot; i++) {
        while(curR > h[i].r) Del(curR--);
        int lst1 = tim1, lst2 = tim2;
        while(curL < h[i].l) Del(curL++);
        ans[h[i].id] = ds.query(h[i].x + 1, n);
        ds.Go(lst1); Go(lst2);
        curL = L;
    }
    ds.Go(n1); Go(n2);
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

    read(n),read(m); B = sqrt(m);
    for(int i = 1, fa; i <= n; i++) {
        cin >> fa;
        if(fa) p[fa].push_back(i);
        else root = i;
    }
    ljm.init(); Prepare();
    for(int i = 1; i <= m; i++) read(z[i].l),read(z[i].r),read(z[i].x),z[i].id = i;
    sort(z + 1, z + m + 1, cmp1);
    int j = 0;
    for(int i = 1; i <= n; i += B) {
        tot = 0;
        while(j + 1 <= m && z[j + 1].l <= i + B - 1) j++, h[++tot] = z[j];
        Sol(i, min(n, i + B - 1)); 
    }
    for(int i = 1; i <= m; i++) print(ans[i]),pc('\n');
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Subtask #1:

score: 0
Runtime Error

Test #1:

score: 0
Runtime Error

input:

7 5
3 1 0 5 3 5 1
1 3 1
5 7 2
1 5 1
4 7 1
4 7 2

output:


result:


Subtask #2:

score: 0
Skipped

Dependency #1:

0%

Subtask #3:

score: 0
Time Limit Exceeded

Test #30:

score: 0
Time Limit Exceeded

input:

50000 200000
42574 43129 47328 17982 40521 6668 12729 32377 201 11940 8599 11734 18349 41045 26854 22540 9897 33419 7463 1243 47272 27135 49050 49111 22435 42539 39924 20272 5843 9308 45963 3283 31185 13692 38952 20583 15885 24802 4773 953 49907 28689 36942 23550 19449 8970 33340 31665 5407 46023 18...

output:


result:


Subtask #4:

score: 0
Skipped

Dependency #3:

0%

Subtask #5:

score: 0
Time Limit Exceeded

Test #67:

score: 0
Time Limit Exceeded

input:

100000 1000000
6457 23693 90928 23592 90440 75018 16865 3342 83718 16731 95103 31510 38719 27886 29093 41955 6596 46409 51839 10527 91993 61074 14405 34833 53674 42363 11490 43757 46191 6058 59164 96938 57858 40178 97523 84164 21582 72243 11267 47368 97058 6637 95208 60092 53943 16441 28363 64965 52...

output:


result:


Subtask #6:

score: 0
Skipped

Dependency #1:

0%