[BZOJ 1691] 挑剔的美食家

[题目链接]

          https://www.lydsy.com/JudgeOnline/problem.php?id=1691

[算法]

         不难想到如下算法 :

         将所有牛和牧草按鲜嫩程度降序排序,按顺序扫描,如果扫描到的是牧草,则将牧草的美味程度加入一个集合,否则,将答案加上比这头牛的期望价格大的牧草中价格最小的

         这个贪心策略的正确性是显然的,具体实现时,我们可以维护一棵平衡树,这棵平衡树支持 : 插入/删除一个数,查询一个数的后继,我们可以方便地使用STL-set完成这个任务,为了练习平衡树,笔者使用的是伸展树

         时间复杂度 : O(Nlog(N))

[代码]

       

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100010

struct info
{
        long long x,y;
        int opt;
} a[MAXN << 1];

int i,n,m;
long long ans,tmp;

template <typename T> inline void read(T &x)
{
        long long 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;
}
struct Splay
{
        int root,total;
        struct Node
        {
                int fa;
                long long val;
                int cnt;
                int son[2];
        }    Tree[MAXN << 1];
        inline bool get(int x)
        {
                return Tree[Tree[x].fa].son[1] == x;
        }
        inline void rotate(int x)
        {
                int f = Tree[x].fa,g = Tree[f].fa;
                int tmpx = get(x),tmpf = get(f);
                if (!f) return;
                Tree[f].son[tmpx] = Tree[x].son[tmpx ^ 1];
                if (Tree[x].son[tmpx ^ 1]) Tree[Tree[x].son[tmpx ^ 1]].fa = f;
                Tree[x].son[tmpx ^ 1] = f;
                Tree[f].fa = x;
                Tree[x].fa = g;
                if (g) Tree[g].son[tmpf] = x;
        }
        inline void splay(int x)
        {
                int fa;
                for (fa = Tree[x].fa; (fa = Tree[x].fa); rotate(x))
                        rotate((get(fa) == get(x)) ? fa : x);
                root = x;
        }
        inline int insert(long long x)
        {
                bool tmp;
                if (!root)
                {
                        root = ++total;
                        Tree[root].fa = 0;
                        Tree[root].son[0] = Tree[root].son[1] = 0;
                        Tree[root].cnt = 1;
                        Tree[root].val = x;
                        return total;
                }
                int now = root;
                while (now > 0)
                {
                        if (Tree[now].val == x)
                        {
                                Tree[now].cnt++;
                                splay(now);
                                return now;
                        }
                        tmp = x > Tree[now].val;
                        if (!Tree[now].son[tmp])
                        {
                                Tree[now].son[tmp] = ++total;
                                Tree[total].fa = now; 
                                Tree[total].val = x;
                                Tree[total].cnt = 1;
                                Tree[total].son[0] = Tree[total].son[1] = 0;
                                splay(total);
                                return total;
                        } else now = Tree[now].son[tmp];
                }
        }
        inline int get_pos(int x)
        {
                int now = x;
                while (Tree[now].son[1]) now = Tree[now].son[1];
                return now;
        }
        inline void join(int x,int y)
        {
                int p = get_pos(x);
                splay(p);
                Tree[p].son[1] = y;
                Tree[y].fa = p;
        }
        inline void erase(int x)
        {
                Tree[x].cnt--;
                if (Tree[x].cnt > 0) return;
                if (!Tree[x].son[0] && !Tree[x].son[1])
                {
                        root = 0;
                        return;
                }
                if (!Tree[x].son[0])
                {
                        root = Tree[x].son[1];
                        Tree[root].fa = 0;
                        return;
                }
                if (!Tree[x].son[1])
                {
                        root = Tree[x].son[0];
                        Tree[root].fa = 0;
                        return;
                }
                join(Tree[x].son[0],Tree[x].son[1]);
        }
        inline long long query(long long x)
        {
                int p = insert(x);
                int now = p;
                if (Tree[p].cnt > 1)
                {
                        erase(p);
                        erase(p);
                        return x;
                }
                now = Tree[p].son[1];
                while (Tree[now].son[0]) now = Tree[now].son[0];
                erase(p);
                long long ret = Tree[now].val;
                splay(now);
                erase(now);
                return ret;
        }
} T;
inline bool cmp(info a,info b)
{
        if (a.y != b.y) return a.y > b.y;
        else return a.opt > b.opt;
}

int main() 
{
        
        read(n); read(m);
        for (i = 1; i <= n; i++)
        {
                read(a[i].x);
                read(a[i].y);
                a[i].opt = 0;
        }
        for (i = 1; i <= m; i++)
        {
                read(a[n + i].x);
                read(a[n + i].y);
                a[n + i].opt = 1;        
        }
        sort(a + 1,a + (n + m) + 1,cmp);
        for (i = 1; i <= n + m; i++)
        {
                if (a[i].opt == 0) 
                {
                        tmp = T.query(a[i].x);
                        if (tmp != -1) 
                        {
                                ans += tmp;
                                continue;
                        }
                        printf("-1\n");
                        return 0;
                } else T.insert(a[i].x);
        }
        printf("%lld\n",ans);
        
        return 0;
    
}

 

posted @ 2018-08-16 22:41  evenbao  阅读(158)  评论(0编辑  收藏  举报