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  • POJ 3667 Hotel(算竞进阶习题)

    线段树区间染色

    题目要求最大的连续段的左端点,我们在查询的时候返回最左端即可,注意查找顺序,应该从左到右!!

    另外这类染色的push_down其实比较简单,直接染成上一层的标记即可

    push_up和连续子段和有点像,需要维护前缀和后缀

    #include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    using namespace std;
    typedef long long ll;
    inline int lowbit(int x){ return x & (-x); }
    inline int read(){
        int X = 0, w = 0; char ch = 0;
        while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
        while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
        return w ? -X : X;
    }
    inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
    inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
    template<typename T>
    inline T max(T x, T y, T z){ return max(max(x, y), z); }
    template<typename T>
    inline T min(T x, T y, T z){ return min(min(x, y), z); }
    template<typename A, typename B, typename C>
    inline A fpow(A x, B p, C yql){
        A ans = 1;
        for(; p; p >>= 1, x = 1LL * x * x % yql)if(p & 1)ans = 1LL * x * ans % yql;
        return ans;
    }
    const int N = 50005;
    int n, m;
    int prefix[N<<2], suffix[N<<2], tree[N<<2], lazy[N<<2];
    
    void push_up(int rt, int l, int r){
        int lson = rt << 1, rson = rt << 1 | 1;
        int mid = (l + r) >> 1;
        tree[rt] = max(tree[lson], tree[rson], suffix[lson] + prefix[rson]);
        prefix[rt] = prefix[lson], suffix[rt] = suffix[rson];
        if(tree[lson] == mid - l + 1) prefix[rt] += prefix[rson];
        if(tree[rson] == r - mid) suffix[rt] += suffix[lson];
    }
    
    void push_down(int rt, int l, int r){
        if(lazy[rt] != -1){
            int lson = rt << 1, rson = rt << 1 | 1;
            int mid = (l + r) >> 1;
            lazy[lson] = lazy[rson] = lazy[rt];
            prefix[lson] = suffix[lson] = tree[lson] = lazy[rt] * (mid - l + 1);
            prefix[rson] = suffix[rson] = tree[rson] = lazy[rt] * (r - mid);
            lazy[rt] = -1;
        }
    }
    
    void buildTree(int rt, int l, int r){
        if(l == r){
            prefix[rt] = suffix[rt] = tree[rt] = 1;
            lazy[rt] = -1;
            return;
        }
        lazy[rt] = -1;
        int mid = (l + r) >> 1;
        buildTree(rt << 1, l, mid);
        buildTree(rt << 1 | 1, mid + 1, r);
        push_up(rt, l, r);
    }
    
    void modify(int rt, int l, int r, int modifyL, int modifyR, int status){
        if(l == modifyL && r == modifyR){
            tree[rt] = prefix[rt] = suffix[rt] = (r - l + 1) * status;
            lazy[rt] = status;
            return;
        }
        push_down(rt, l, r);
        int mid = (l + r) >> 1;
        if(modifyL > mid) modify(rt << 1 | 1, mid + 1, r, modifyL, modifyR, status);
        else if(modifyR <= mid) modify(rt << 1, l, mid, modifyL, modifyR, status);
        else{
            modify(rt << 1, l, mid, modifyL, mid, status);
            modify(rt << 1 | 1, mid + 1, r, mid + 1, modifyR, status);
        }
        push_up(rt, l, r);
    }
    
    int query(int rt, int l, int r, int len){
        if(prefix[rt] >= len) return l;
        push_down(rt, l, r);
        int mid = (l + r) >> 1, lson = rt << 1, rson = rt << 1 | 1;
        if(tree[lson] >= len) return query(lson, l, mid, len);
        //else if(tree[rson] >= len) return query(rson, mid + 1, r, len);
        else if(suffix[lson] + prefix[rson] >= len) return mid - suffix[lson] + 1;
        else if(tree[rson] >= len) return query(rson, mid + 1, r, len);
        else return 0;
    }
    
    int main(){
    
        n = read(), m = read();
        buildTree(1, 1, n);
        while(m --){
            int opt = read();
            if(opt == 1){
                int len = read();
                int pos = query(1, 1, n, len);
                if(pos != 0 ) modify(1, 1, n, pos, pos + len - 1, 0);
                printf("%d
    ", pos);
            }
            else if(opt == 2){
                int x = read(), y = read();
                modify(1, 1, n, x, x + y - 1, 1);
            }
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/onionQAQ/p/10533486.html
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