bzoj1227

Posted 2020-10-02 123456

离散化+树状数组+排列组合

很久以前就看到过这道题，现在依然不会做。。。看完题解发现思路很简单，就是有点难写

我们先将坐标离散化，x和y最大是w，然后我们就有了一个暴力做法， 枚举每块墓地，统计，因为墓地上下左右没东西的话就不可能有贡献，这些坐标自然就被离散化了，所以墓地最多有w*w块

复杂度O(w*w)，然后我们优化一下，发现不用枚举每个墓地，对于相邻两棵树，他们中间墓地左右的树的数量是相同的，这样的方案有C(l,k)*C(r,k)，对于每个墓地，这样的虔诚度是C(l,k)*C(r,k)*C(up,k)*C(down,k),我们发现两棵树中间的墓地C(l,k)*C(r,k)是相同的，现在需要的就是统计sigma(C(up,k)*C(down,k)),我们可以用树状数组维护这个东西，先把墓地按行排序分类，然后每次对于每行枚举相邻的两棵树，计算之间的sigma，这样用树状数组统计，然后就是更新，有树的地方是需要更新的，就是把对应的y坐标离散化后的位置在树状数组里更新，减去原来的值，改成现在新的C(up,k)*C(down,k)

#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
const long long mod = 2147483648ll;
int n, w, m, k;
int up[N], down[N];
long long c[N][12], BIT[N], ans;
struct Tree {
    int x, y, ux, uy, left, right, up, down;
    bool friend operator < (Tree A, Tree B) 
    {
        return A.ux == B.ux ? A.uy < B.uy : A.ux < B.ux;
    }
} tree[N];
vector<Tree> line[N];
inline int lowbit(int i) 
{
    return i & (-i);
}
inline void update(int pos, int delta)
{
    for(int i = pos; i <= w + 5; i += lowbit(i)) BIT[i] = (BIT[i] + delta) % mod;
}
inline long long query(int pos)
{
    long long ret = 0;
    for(int i = pos; i; i -= lowbit(i)) ret = (ret + BIT[i]) % mod;
    return ret;
}
int main()
{
    scanf("%d%d%d", &n, &m, &w);
    vector<int> vx, vy;
    for(int i = 1; i <= w; ++i) 
    {
        scanf("%d%d", &tree[i].x, &tree[i].y);
        vx.push_back(tree[i].x);
        vy.push_back(tree[i].y);
    }
    scanf("%d", &k);
    c[0][0] = 1;
    for(int i = 1; i <= w; ++i)
    {
        c[i][0] = 1;
        for(int j = 1; j <= min(k, i); ++j) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
    sort(vx.begin(), vx.end());
    sort(vy.begin(), vy.end());
    vx.erase(unique(vx.begin(), vx.end()), vx.end());
    vy.erase(unique(vy.begin(), vy.end()), vy.end());   
    for(int i = 1; i <= w; ++i)
    {
        tree[i].ux = lower_bound(vx.begin(), vx.end(), tree[i].x) - vx.begin() + 1;
        tree[i].uy = lower_bound(vy.begin(), vy.end(), tree[i].y) - vy.begin() + 1; 
//      printf("ux=%d uy=%d\n", tree[i].ux, tree[i].uy);
    }
    sort(tree + 1, tree + w + 1);
    for(int i = 1; i <= w; ++i)
    {
        ++up[tree[i].uy];
        tree[i].up = up[tree[i].uy];        
    }
    for(int i = w; i; --i) 
    {
        tree[i].down = down[tree[i].uy];
        ++down[tree[i].uy]; 
    }
    for(int i = 1; i <= w; ++i) line[tree[i].ux].push_back(tree[i]);
    for(int i = 1; i <= vx.size(); ++i) 
    {
        sort(line[i].begin(), line[i].end());   
        for(int j = 0; j < line[i].size(); ++j)
        {
            Tree &tmp = line[i][j];
            tmp.left = j + 1;
            tmp.right = line[i].size() - j; 
        }
    }
    for(int i = 1; i <= vx.size(); ++i)
    {
        for(int j = 1; j < line[i].size(); ++j)
        {
            Tree tmp_l = line[i][j - 1], tmp_r = line[i][j];
            long long comb_left = c[tmp_l.left][k], comb_right = c[tmp_r.right][k], sigma = query(tmp_r.uy - 1) - query(tmp_l.uy);
            ans = (ans + comb_left % mod * comb_right % mod * sigma % mod) % mod;       
        }
        for(int j = 0; j < line[i].size(); ++j)
        {
            Tree tmp = line[i][j];
            update(tmp.uy, c[tmp.up][k] % mod * c[tmp.down][k] % mod - query(tmp.uy) + query(tmp.uy - 1));
        }
    }
    printf("%lld\n", (ans % mod + mod) % mod);
    return 0;
}

View Code