• CodeForces121E 线段树上线段果


    http://codeforces.com/problemset/problem/121/E

    题意: Petya 喜欢幸运数,幸运数只包含 4 和 7 这两个数字。例如 47,744,4 都是幸运数字,但 5,16,467 不是。

    Petya 有一个 N 个数的数组,他想给这个数组执行 M 个操作,可以分为两种操作:

    1. add l r d 把第 l 到 第 r 个数都加上 d; 
    2. count l r 统计第 l 到第 r 个数有多少个幸运数字。

    喜闻乐见的数据结构题。

    更加喜闻乐见的是这题能用树状数组直接暴力跑过去。对每一个区间修改进行n次单点修改的傻狗操作竟然也可以AC这题

    #include <map>
    #include <set>
    #include <ctime>
    #include <cmath>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <string>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <sstream>
    #include <iostream>
    #include <algorithm>
    #include <functional>
    using namespace std;
    #define For(i, x, y) for(int i=x;i<=y;i++)  
    #define _For(i, x, y) for(int i=x;i>=y;i--)
    #define Mem(f, x) memset(f,x,sizeof(f))  
    #define Sca(x) scanf("%d", &x)
    #define Scl(x) scanf("%lld",&x);  
    #define Pri(x) printf("%d
    ", x)
    #define Prl(x) printf("%lld
    ",x);  
    #define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
    #define LL long long
    #define ULL unsigned long long  
    #define mp make_pair
    #define PII pair<int,int>
    #define PIL pair<int,long long>
    #define PLL pair<long long,long long>
    #define pb push_back
    #define fi first
    #define se second 
    typedef vector<int> VI;
    const double eps = 1e-9;
    const int maxn = 1e5 + 10;
    const int INF = 0x3f3f3f3f;
    const int mod = 1e9 + 7; 
    int N,M,tmp,K;
    const int sp[30] = {4,7,44,47,74,77,444,447,474,477,744,747,774,777,4444,4447,4474,4477,4744,4747,4774,4777,7444,7447,7474,7477,7744,7747,7774,7777,};
    bool check[10010];
    int a[maxn];
    int tree[maxn];
    void add(int t,int x){
        for(;t <= N;t += t & -t){
            tree[t] += x;
        }
    }
    int getsum(int t){
        int s = 0;
        for(;t;t -= t & -t){
            s += tree[t];
        }
        return s;
    }
    int main()
    {
        For(i,0,29) check[sp[i]] = 1;
        scanf("%d%d",&N,&M);
        For(i,1,N){
            int x; Sca(x); a[i] = x;
            if(check[x]) add(i,1);
        }
        For(i,1,M){
            int l,r;
            char op[20];
            scanf("%s%d%d",op,&l,&r);
            if(op[0] == 'a'){
                int d; Sca(d);
                For(j,l,r){
                    if(check[a[j]]) add(j,-1);
                    a[j] += d;
                    if(check[a[j]]) add(j,1);
                }
            }else{
                Pri(getsum(r) - getsum(l - 1));
            }
        }
        #ifdef VSCode
        system("pause");
        #endif
        return 0;
    }
    View Code

    当然正解肯定不是这样的

    困难之处在于维护区间内幸运数字的个数。事实上确实不简单,思路和hdu多校里面的一道线段树相似。

    线段树维护三个值,区间内所有数到下一个幸运数字差值的最小值,最小值的个数,lazy标记

    每一次update之后要重新遍历一次线段树去寻找有没有最小值小于0的数,然后将这些减少到0的数一个个修改即可。

    看起来很麻烦的操作事实上时间复杂度是很科学的正解。

    #include <map>
    #include <set>
    #include <ctime>
    #include <cmath>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <string>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <sstream>
    #include <iostream>
    #include <algorithm>
    #include <functional>
    using namespace std;
    #define For(i, x, y) for(int i=x;i<=y;i++)  
    #define _For(i, x, y) for(int i=x;i>=y;i--)
    #define Mem(f, x) memset(f,x,sizeof(f))  
    #define Sca(x) scanf("%d", &x)
    #define Scl(x) scanf("%lld",&x);  
    #define Pri(x) printf("%d
    ", x)
    #define Prl(x) printf("%lld
    ",x);  
    #define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
    #define LL long long
    #define ULL unsigned long long  
    #define mp make_pair
    #define PII pair<int,int>
    #define PIL pair<int,long long>
    #define PLL pair<long long,long long>
    #define pb push_back
    #define fi first
    #define se second 
    inline LL read(){LL now=0;register char c=getchar();for(;!isdigit(c);c=getchar());
    for(;isdigit(c);now=now*10+c-'0',c=getchar());return now;}
    inline void out(int x){if(x > 9) out(x / 10); putchar(x % 10 + '0');}
    inline void out(LL x){if(x > 9) out(x / 10); putchar(x % 10 + '0');}
    typedef vector<int> VI;
    const double eps = 1e-9;
    const int maxn = 1e5 + 10;
    const int INF = 0x3f3f3f3f;
    const int mod = 1e9 + 7; 
    int N,M,tmp,K;
    const int sp[31] = {4,7,44,47,74,77,444,447,474,477,744,747,774,777,4444,4447,4474,4477,4744,4747,4774,4777,7444,7447,7474,7477,7744,7747,7774,7777,44444};
    struct Tree{
        int l,r;
        int x,y,lazy;
    }tree[maxn * 4];
    int c[maxn];
    int a[maxn];
    void Pushup(int t){
        if(tree[t << 1].x == tree[t << 1 | 1].x){
            tree[t].x = tree[t << 1].x;
            tree[t].y = tree[t << 1].y + tree[t << 1 | 1].y;
        }else if(tree[t << 1].x < tree[t << 1 | 1].x){
            tree[t].x = tree[t << 1].x;tree[t].y = tree[t << 1].y;
        }else{
            tree[t].x = tree[t << 1 | 1].x;tree[t].y = tree[t << 1 | 1].y;
        }
    }
    void Build(int t,int l,int r){
        tree[t].l = l; tree[t].r = r;
        tree[t].lazy = 0;
        if(l == r){
            tree[t].x = c[a[l]];
            tree[t].y = 1;
            return;
        }
        int m = (l + r) >> 1;
        Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r);
        Pushup(t);
    }
    void Pushdown(int t){
        if(tree[t].lazy){
            tree[t << 1].lazy += tree[t].lazy;
            tree[t << 1 | 1].lazy += tree[t].lazy;
            tree[t << 1].x += tree[t].lazy;
            tree[t << 1 | 1].x += tree[t].lazy;
            tree[t].lazy = 0;
        }
    }
    void update(int t,int l,int r,int val){
        if(l <= tree[t].l && tree[t].r <= r){
            tree[t].x += val;
            tree[t].lazy += val;
            return;
        }
        Pushdown(t);
        int m = (tree[t].l + tree[t].r) >> 1;
        if(r <= m) update(t << 1,l,r,val);
        else if(l > m) update(t << 1 | 1,l,r,val);
        else{
            update(t << 1,l,m,val); update(t << 1 | 1,m + 1,r,val);
        }
        Pushup(t);
    }
    
    void rebuild(int t){
        if(tree[t].x >= 0) return;
        if(tree[t].l == tree[t].r){
            a[tree[t].l] -= tree[t].lazy;
            tree[t].x = c[a[tree[t].l]];
            tree[t].lazy = 0;
            return ;
        }
        Pushdown(t);
        rebuild(t << 1); rebuild(t << 1 | 1);
        Pushup(t);
    }
    int query(int t,int l,int r){
        if(tree[t].x) return 0;
        if(l <= tree[t].l && tree[t].r <= r){
            return tree[t].y;
        }
        Pushdown(t);
        int m = (tree[t].l + tree[t].r) >> 1;
        if(r <= m) return query(t << 1,l,r);
        if(l > m) return query(t << 1 | 1,l,r);
        else return query(t << 1,l,m) + query(t << 1 | 1,m + 1,r);
    }
    int main()
    {
        int cnt = 0;
        For(i,1,1e4){
            if(sp[cnt] < i) cnt++;
            c[i] = sp[cnt] - i;
        }
        N = read(); M = read();
        For(i,1,N) a[i] = read();
        Build(1,1,N);
        int l,r;
        char op[20];
        For(i,1,M){
            scanf("%s",op);
            l = read(); r = read();
            if(op[0] == 'a'){
                int d; d = read();
                update(1,l,r,-d);
                rebuild(1);
            }else{
                out(query(1,l,r));
                puts("");
            }
        }
        #ifdef VSCode
        system("pause");
        #endif
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Hugh-Locke/p/9598743.html
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