题目链接：【http://acm.split.hdu.edu.cn/showproblem.php?pid=5517】
题意：定义multi_set A<a , d>，B<c , d , e>，C<x , y , z>，给出 A , B ，定义 C = A * B = ={⟨a,c,d⟩∣⟨a,b⟩∈A, ⟨c,d,e⟩∈B and b=e} 。求出C之后，求C中一个元素t[i]<a , b , c>是否存在一个元素tmp<x , y , z>使得 x >= a&&y > =b&&z >= c && t != tmp;如果不存在ans++;输出ans即可。
题解：
首先，我们要剪枝和去重，加入集合A中存在<1 , 3>,<2, 3>,<3, 3> ,<4, 3>，<4, 3>，那么只需要保留<4,3>就可以了，并且记录一下数量。因为前面的数都可以找到比它大的元素。求出C之后，我们依旧要对C去重。那么这道题就是求C某个元素是否存在大于它的元素了。因为这道题中B set里面c,d,很小在生成的C中y,z也很小，都小于1000,那么我们就可以用二维树状数组做了。首先对C进行从大到小排序，树状数组中只记录y,z的值。我们枚举C中的额每一个元素，先查询(1001 - y ,1001 - z)，然后在更新（1001 - y ,1001 - z）,应为是从大到小排序的，所以后面的元素的x值一定小于等于前面的元素的x值，s如果查询不为0,那么一定存在某个元素是大于它的。我们一可以直接套CDQ分治的三维偏序模板，为维护值即可。
树状数组：
#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
", cs, ans);
}
return 0;
}
三维偏序：
#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 < b.y;
}
//--------------------------------------------------------//树状数组
int c[maxn], maxz = 1001;
inline int lowbit(int x)
{
return 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
", ++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
", ic, ret);
}
return 0;
}