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  • 【USACO2017JAN】 Promotion Counting

    【题目链接】

                点击打开链接

    【算法】

              离散化 + dfs + 树状数组

    【代码】

              

    #include<bits/stdc++.h>
    using namespace std;
    #define MAXN 100000
    
    int N,i,lth,x,pos;
    int a[MAXN+10],tmp[MAXN+10],rank[MAXN+10],ans[MAXN+10];
    vector<int> E[MAXN+10];
    
    template <typename T> inline void read(T &x) {
            int f = 1; x = 0;
            char c = getchar();
            for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
            for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
            x *= f;
    }
    
    template <typename T> inline void write(T x) {
        if (x < 0) { putchar('-'); x = -x; }
        if (x > 9) write(x/10);
        putchar(x%10+'0');
    }
    
    template <typename T> inline void writeln(T x) {
        write(x);
        puts("");
    }
    
    struct BinaryIndexedTree {
            int bit[MAXN+10];
            int lowbit(int x) { return x & -x; }
            inline void modify(int pos) {
                    int i;
                    for (i = pos; i <= N; i += lowbit(i)) bit[i]++;
            }
            inline int query(int pos) {
                    int i,ret = 0;
                    for (i = pos; i; i -= lowbit(i)) ret += bit[i];
                    return ret;
            }
    } BIT;
    
    inline void dfs(int x) {
            int i;
            BIT.modify(rank[x]);
            ans[x] = -BIT.query(rank[x]-1);
            for (i = 0; i < E[x].size(); i++) dfs(E[x][i]);
            ans[x] += BIT.query(rank[x]-1);
    }
    
    int main() {
        
            read(N);
            for (i = 1; i <= N; i++) {
                    read(a[i]);
                    tmp[i] = a[i];
            }
            
            sort(tmp+1,tmp+N+1);
            for (i = 1; i <= N; i++) {
                    if (tmp[i] != tmp[i-1]) 
                            tmp[++lth] = tmp[i];
            }
            
            for (i = 1; i <= N; i++) {
                    pos = lower_bound(tmp+1,tmp+lth+1,a[i]) - tmp;
                    rank[i] = N - pos + 1;
            }
            
            for (i = 2; i <= N; i++) {
                    read(x);
                    E[x].push_back(i);
            }
            
            dfs(1);
            
            for (i = 1; i <= N; i++) writeln(ans[i]);
            
             return 0;
        
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/9196384.html
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