Hacker Cups and Balls Gym - 101234A 二分+线段树

题目:题目链接

题意:有编号从1到n的n个球和n个杯子. 每一个杯子里有一个球, 进行m次排序操作，每次操作给出l,r. 如果l<r，将[l,r]范围内的球按升序排序, 否则降序排, 问中间位置的数是多少.

思路:

　　暴力复杂度为m*nlog(n), 不能暴力排序

　　二分答案, 对于当前mid, 我们将大于等于mid的数记为1, 否则记0, 则排序就是将某个区间的数改为1或0, 通过线段树区间更新可以方便的做到, 对排序后的结果查询判断二分区间应该取左还是取右, 若中间的数是1, 则说明答案大于等于当前的数, l右移, 否则左移

AC代码:

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <map>
 #include <set>
 #include <unordered_map>
 #include <unordered_set>
 #include <queue>
 #include <stack>
 #include <list>

 #define FRER() freopen("in.txt", "r", stdin)
 #define FREW() freopen("out.txt", "w", stdout)

 #define INF 0x3f3f3f3f
 #define eps 1e-8

 using namespace std;

 const int maxn = 1e5 + ;

 int a[maxn], ql[maxn], qr[maxn], tree[maxn << ], lazy[maxn << ];

 void build(int l, int r, int rt, int num) {
     lazy[rt] = ;
     if(l == r) {
         tree[rt] = (a[l] >= num);
         return ;
     }
     int m = (l + r) >> ;
     build(l, m, rt << , num);
     build(m + , r, rt << |, num);
     tree[rt] = tree[rt << ] + tree[rt << |];
 }

 void pushdown(int l, int r, int rt) {
     int m = (l + r) >> ;
     lazy[rt << ] = lazy[rt << |] = lazy[rt];
     if(lazy[rt] == ) {
         tree[rt << |] = min(tree[rt], r - m);
         tree[rt << ] = tree[rt] - tree[rt << |];
     }
     else if(lazy[rt] == ){
         tree[rt << ] = min(tree[rt], m - l + );
         tree[rt << |] = tree[rt] - tree[rt << ];
     }
     lazy[rt] = ;
 }

 int querySum(int x, int y, int l, int r, int rt) {
     if(x <= l && y >= r) return tree[rt];
     if(lazy[rt]) pushdown(l, r, rt);
     int m = (l + r) >> , sum = ;
     if(x <= m) sum += querySum(x, y, l, m, rt << );
     if(y > m) sum += querySum(x, y, m + , r, rt << |);
     return sum;
 }

 int query(int pos, int l, int r, int rt) {
     if(l == r) return tree[rt];
     if(lazy[rt]) pushdown(l, r, rt);
     int m = (l + r) >> ;
     if(pos <= m) return query(pos, l, m, rt << );
     else return query(pos, m + , r, rt << |);
 }

 void update(int x, int y, int l, int r, int rt, int flag, int sum) {
     if(x <= l && y >= r) {
         tree[rt] = sum;
         lazy[rt] = flag;
         return ;
     }
     if(lazy[rt]) pushdown(l, r, rt);
     int m = (l + r) >> ;
     if(y <= m) update(x, y, l, m, rt << , flag, sum);
     else if(x > m) update(x, y, m + , r, rt << |, flag, sum);
     else {
         int lsum, rsum;
         if(flag== ) {
             rsum = min(sum, y - m);
             lsum = sum - rsum;
         }
         else if(flag == ){
             lsum = min(sum, m - x + );
             rsum = sum - lsum;
         }
         update(x, m, l, m, rt << , flag, lsum);
         update(m + , y, m + , r, rt << |, flag, rsum);
     }
     tree[rt] = tree[rt << ] + tree[rt << |];
 }

 bool judge(int n, int m, int mid) {
     build(, n, , mid);
     for(int i = ; i < m; ++i) {
         int sum = querySum(min(ql[i], qr[i]), max(ql[i], qr[i]), , n, );
         if(ql[i] <= qr[i]) update(ql[i], qr[i], , n, , , sum);
         else update(qr[i], ql[i], , n, , , sum);
     }
     return query(( + n) >> , , n, );
 }

 int main()
 {
     //FRER();
     ios::sync_with_stdio();
     cin.tie();

     int n, m;
     cin >> n >> m;
     for(int i = ; i <= n; ++i) cin >> a[i];
     for(int i = ; i < m; ++i) cin >> ql[i] >> qr[i];
     int l = , r = n, ans;
     while(l <= r) {
         int mid = (l + r) >> ;
         if(judge(n, m, mid))
             ans = mid, l = mid + ;
         else r = mid - ;
     }
     cout << ans << endl;
     return ;
 }