zoukankan      html  css  js  c++  java
  • P4269 [USACO18FEB]Snow Boots G

    思维题。 以地板为序构造链表,再排序,然后删除走不过去的地面。 删除的时候顺便维护最大的跨度,以此判断可行性。 总的来说利用了答案的单调性。
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    const int MAXN = 1e5 + 20;
    inline int read()
    {
        int x = 0; char ch = getchar(); bool f = false;
        while(!isdigit(ch)) f |= (ch == '-'), ch = getchar();
        while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
        return f ? -x : x;
    }
    
    int N, B;
    struct Node
    {
        int val;
        Node *pre, *nxt;
    }node[MAXN];
    
    struct boot
    {
        int dep, dis, idx, ans;
        inline bool operator >(const boot &rhs) const {
            return dep > rhs.dep;
        }
        inline bool operator <(const boot &rhs) const {
            return idx < rhs.idx;
        }
    }b[MAXN];
    struct Floor
    {
        int dep, idx;
        Node *pos;
        inline bool operator >(const Floor &rhs) const {
            return dep > rhs.dep;
        }
        inline bool operator <(const Floor &rhs) const {
            return idx < rhs.idx;
        }
    }f[MAXN];
    
    
    void build(){
        f[1].pos = &node[1]; node[1].val = 1;
        for(int i = 2; i <= N; i++){
            node[i].pre = &node[i - 1], node[i - 1].nxt = &node[i];
            node[i].val = 1, f[i].pos = &node[i];
        }
    }
    
    int main()
    {
        cin>>N>>B;
        for(int i = 1; i <= N; i++) f[i] = (Floor){read(), i};
        for(int i = 1; i <= B; i++) 
            b[i].dep = read(), b[i].dis = read(), b[i].idx = i;
        
        build();
        sort(f + 1, f + N + 1, greater<Floor>());
        sort(b + 1, b + B + 1, greater<boot>());
    
        int p = 1, maxs = 1;
        for(int i = 1; i <= B; i++){
            while(p <= N && f[p].dep > b[i].dep) {
                Node *cur = f[p].pos;
                cur->pre->nxt = cur->nxt;
                cur->nxt->pre = cur->pre;
                maxs = max(maxs, cur->pre->val += cur->val);
                ++p;
            }
            if(maxs > b[i].dis) b[i].ans = 0;
            else b[i].ans = 1;
        }
        sort(b + 1, b + B + 1);
        for(int i = 1; i <= B; i++) printf("%d
    ", b[i].ans);
        return 0;
    }
    
    
  • 相关阅读:
    完美解决php无法上传大文件分享
    完美解决php无法上传大文件问题
    完美解决php无法上传大文件思路
    完美解决php无法上传大文件功能
    IfcRightCircularCylinder
    IfcRightCircularCone
    IfcRectangularPyramid
    IfcBlock
    IfcCsgPrimitive3D
    IfcReparametrisedCompositeCurveSegment
  • 原文地址:https://www.cnblogs.com/wsmrxc/p/9439965.html
Copyright © 2011-2022 走看看