HDU 5517 "Two-dimensional tree-like array///three-dimensional partial order problem"

Topic Link: "http://acm.split.hdu.edu.cn/showproblem.php?pid=5517"

Test instructions: DefiningMulti_set A<a, D>,b<c, D, E>,c<x, Y, z>, give a, B, define C = A * B =={?A,C,D?∣?A,B?∈a,  ? C,d ,e ? ∈b  and b=e< Span id= "mathjax-span-63" class= "Mo" >} 。after finding C, ask for an element in C T[i]<a, B, c> if there is an element tmp<x, Y, z> make x >= a&&y > =b&&z >= C & & T! = tmp; If ans++ is not present, output ans.

Exercises

First of all, we want to prune and go to the weight, join the set a in the presence of <1, 3>,<2, 3>,<3, 3>, <4, 3>,<4, 3> then just keep <4,3> on it, and record the number. Because the previous number can find an element larger than it. After we find C, we still have to go to the C-weight. So the problem is to ask whether an element of C is larger than its element. Because this problem in the B set inside c,d, very small in the generated C y,z is also very small, are less than 1000, then we can use a two-dimensional tree array to do. First, C is sorted from large to small, and only y,z values are recorded in the tree array. We enumerate the amount of each element in C, first query (1001-y, 1001-z), and then in the update (1001-Y, 1001-z), should be from large to small sort, so the next element of the X value must be less than equal to the previous element's X value, s if the query is not 0, then there must be an element is Larger than it. We can directly set CDQ division of the three-dimensional partial-order template, in order to maintain the value.

Tree-like array:

#include <bits/stdc++.h>using namespace Std;typedef long long ll;const int MAXN = 1e5 + 15;int ic, N, M;struct point{    int x, y, Z;    LL W;     Point () {} point (int x, int y, int z, LL W): X (x), Y (y), Z (z), W (w) {} BOOL operator < (const point& T) const        {if (x! = t.x) return T.x < x;        if (Y! = t.y) return T.y < Y;    return t.z < Z;    } bool operator = = (Const point T) Const {return (x = = T.x && y = = T.y && z = = t.z); }} p[maxn];int T[maxn], c[maxn];int tbit[1050][1050];int low_bit (int x) {return x & x;} void Bit_add (int x, int y, int val) {for (int i = x; i <= 1001; i + = Low_bit (i)) for (int j = y; J <= 1001; J + = Low_bit (j)) Tbit[i][j] + = val;    int bit_sum (int x, int y) {int ret = 0;    for (int i = x; i; i-= Low_bit (i)) for (int j = y; j; J-= Low_bit (j)) ret + = Tbit[i][j]; return ret;}    int main () {scanf ("%d", &ic); for (int cs = 1; CS <= IC;        cs++) {scanf ("%d%d", &n, &m);        memset (t, 0, sizeof (t));            for (int i = 1; I <= N; i++) {int A, B;            scanf ("%d%d", &a, &b);            if (a > T[b]) t[b] = A, c[b] = 1;        else if (a = = T[b]) c[b]++;        } int tmp = 0;            for (int i = 1; I <= M; i++) {int A, b, C;            scanf ("%d%d%d", &a, &b, &c);        if (T[c]) p[++tmp] = Point (T[c], A, B, (LL) c[c]);        } sort (p + 1, p + 1 + tmp);        N = 1;            for (int i = 2; I <= tmp; i++) {if (p[i] = = P[n]) P[N].W + = P[I].W;        else p[++n] = p[i];        } LL ans = 0;        memset (tbit, 0, sizeof (tbit)); for (int i = 1; I <= N; i++) {if (!            Bit_sum (1001-P[I].Y, 1001-p[i].z)) ans + = P[I].W;        Bit_add (1001-p[i].y, 1001-p[i].z, 1);    } printf ("Case #%d:%lld\n", cs, ans); } return 0;} 

three-dimensional partial order:

#include <bits/stdc++.h>using namespace Std;typedef long long ll;const int MAXN = 1e5 + 15;int cs, N, M;int T[MAXN], C[MAXN];    LL W[MAXN], ans[maxn];struct edge{int x, y, z, id;    LL W; Edge () {} edge (int x, int y, int z, int id, LL W): X (x), Y (y), Z (z), ID (ID), W (w) {} BOOL operator < (const Edge    & T) const {return Z < t.z;    } bool operator = = (Const Edge & T) const {return t.x = = x && t.y = = y && t.z = = Z;  }} B[MAXN], P[MAXN], T[maxn];bool cmpx (Edge &a, Edge &b) {if (a.x = b.x && a.y = b.y && A.z = =    B.Z) return a.id < b.id;    else if (a.x = = b.x && a.y = = b.y) return a.z < b.z;    else if (a.x = = b.x) return A.Y < b.y; else return a.x < b.x;}    BOOL Cmpy (Edge &a, Edge &b) {if (a.y = = B.y && a.z = = b.z) return a.id < b.id;    else if (a.y = = b.y) return a.z < b.z; else return A.Y &LT B.Y;} --------------------------------------------------------//tree array int C[MAXN], maxz = 1001;inline int lowbit (int x) {RET Urn x & (-X);} void Add (int x, int val) {while (x <= maxz) c[x] + = val, x + = Lowbit (x);}    int get_sum (int x) {int ret = 0;    while (x > 0) ret + = C[x], X-= Lowbit (x); return ret;}    -------------------------------------------------void Cdq (int l, int r) {if (L = = r) return;    int mid = (L + r)/2;    CDQ (L, mid), CDQ (mid + 1, R);    for (int i = l; I <= R; i++) T[i] = P[i];    Sort (T + L, T + mid + 1, cmpy), sort (T + mid + 1, T + R + 1, cmpy);    int a1 = L, a2 = mid + 1;        for (int i = l; I <= R; i++) {if (A1 = = mid + 1) ans[t[a2].id] + = Get_sum (t[a2].z), a2++;        else if (A2 = = R + 1) Add (t[a1].z, 1), a1++;        else if (t[a1].y <= t[a2].y) Add (t[a1].z, 1), a1++;    else Ans[t[a2].id] + = Get_sum (t[a2].z), a2++; } for (int i = l; I <= mid; i++) Add (t[i].z,-1);}int main () {scanf ("%d", &cs);        for (int ic = 1; IC <= CS; ic++) {memset (t, 0, sizeof (t));        memset (ans, 0, sizeof (ans));        scanf ("%d%d", &n, &m);            for (int i = 1; I <= N; i++) {int A, B;            scanf ("%d%d", &a, &b);            if (a > T[b]) t[b] = A, c[b] = 1;        else if (a = = T[b]) c[b]++;        } int tmp = 0;            for (int i = 1; I <= M; i++) {int A, b, C;            scanf ("%d%d%d", &a, &b, &c);        if (T[c]) ++tmp, p[tmp] = Edge (T[c], a, B, I, (LL) c[c]);        } sort (p + 1, p + 1 + tmp, CMPX);        N = 1;            for (int i = 2; I <= tmp; i++) {if (p[i] = = P[n]) P[N].W + = P[I].W;        else p[++n] = p[i];            } if (N <= 1) {printf ("Case #%d:0\n", ++ic);        Continue            } for (int i = 1; I <= N; i++) {p[i].x = 100001-p[i].x;P[i].y = 1001-P[I].Y;        P[i].z = 1001-p[i].z;            } for (int i = 1; I <= N; i++) {p[i].id = i;        W[i] = P[I].W;        } sort (p + 1, p + 1 + N, cmpx);        CDQ (1, N);        P[n + 1].x =-1; for (int i = N; I >= 1; i--) {if (p[i].x = = p[i + 1].x && p[i].y = p[i + 1].y && P[i]        . z = = p[i + 1].z) ans[p[i].id] = ans[p[i + 1].id];        } LL ret = 0;        for (int i = 1; I <= N; i++) if (!ans[i]) ret + = W[i];    printf ("Case #%d:%lld\n", IC, ret); } return 0;}

　　

HDU 5517 "Two-dimensional tree-like array///three-dimensional partial order problem"