G. Trace ACM-ICPC 2018 徐州赛区网络预赛线段树写法

There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy ) means the wave is a rectangle whose vertexes are ( 00 , 00 ), ( xx , 00 ), ( 00 , yy ), ( xx , yy ). Every time the wave will wash out the trace of former wave in its range and remain its own trace of ( xx , 00 ) -> ( xx , yy ) and ( 00 , yy ) -> ( xx , yy ). Now the toad on the coast wants to know the total length of trace on the coast after n waves. It's guaranteed that a wave will not cover the other completely.

Input

The first line is the number of waves n(n \le 50000)n(n≤50000).

The next nn lines，each contains two numbers xx yy ,( 0 < x0<x , y \le 10000000y≤10000000 )，the ii-th line means the ii-th second there comes a wave of ( xx , yy ), it's guaranteed that when 1 \le i1≤i , j \le nj≤n ，x_i \le x_jxi≤xj and y_i \le y_jyi≤yj don't set up at the same time.

Output

An Integer stands for the answer.

Hint：

As for the sample input, the answer is 3+3+1+1+1+1=103+3+1+1+1+1=10

样例输入复制

样例输出复制

　　　第一眼扫描线想想突然感觉不太对，扫描线不会写

　　　　这题仔细分析一下这题倒推更加容易写

　　　　这题倒推的话只要更新比这个矩形更大的区域就行了

　　　　离散化一下

　　　　注意两个点

　　　　　　　int ny = query( 1, L[i], tot, 1 );
　　　　　　　int nx = query( 2, R[i], tot, 1 );

　　　　因为他给的是一个矩形所以我们看这条边能不能算就是看这个点的左边的最大值所以query （L[I]->tot）

　　　　因为你只要左边有一条边，这是一个矩形所以右边也一定有

　　　　第二个点

　　　　　　update( 1, L[i], sum[R[i] - 1], 1 );
　　　　　　update( 2, R[i], sum[L[i] - 1], 1 );

　　　　L[i] 表示X轴上的点他的值对应的是 sum[R[i] - 1]

 #include <cstdio>

 #include <cstring>

 #include <queue>

 #include <cmath>

 #include <algorithm>

 #include <set>

 #include <iostream>

 #include <map>

 #include <stack>

 #include <string>

 #include <vector>

 #define  pi acos(-1.0)

 #define  eps 1e-6

 #define  fi first

 #define  se second

 #define  lson l,m,rt<<1

 #define  rson m+1,r,rt<<1|1

 #define  rtl   rt<<1

 #define  rtr   rt<<1|1

 #define  bug         printf("******\n")

 #define  mem(a,b)    memset(a,b,sizeof(a))

 #define  name2str(x) #x

 #define  fuck(x)     cout<<#x" = "<<x<<endl

 #define  f(a)        a*a

 #define  sf(n)       scanf("%d", &n)

 #define  sff(a,b)    scanf("%d %d", &a, &b)

 #define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)

 #define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)

 #define  pf          printf

 #define  FRE(i,a,b)  for(i = a; i <= b; i++)

 #define  FREE(i,a,b) for(i = a; i >= b; i--)

 #define  FRL(i,a,b)  for(i = a; i < b; i++)

 #define  FRLL(i,a,b) for(i = a; i > b; i--)

 #define  FIN         freopen("in.txt","r",stdin)

 #define  gcd(a,b)    __gcd(a,b)

 #define  lowbit(x)   x&-x

 using namespace std;

 typedef long long  LL;

 typedef unsigned long long ULL;

 const int INF = 0x7fffffff;

 const int mod = 1e9 + ;

 const int maxn = 1e6 + ;

 struct node {

     int l, r, sumx, sumy;

     int mid() {

         return ( l + r ) >>  ;

     }

 } tree[maxn << ];

 void build( int l, int r, int rt ) {

     tree[rt].l = l, tree[rt].r = r, tree[rt].sumx = tree[rt].sumy = ;

     if ( l == r ) return ;

     int m = ( l + r ) >> ;

     build( l, m, rtl );

     build( m + , r, rtr );

 }

 void update( int op, int pos, int key, int rt ) {

     if ( tree[rt].l == tree[rt].r ) {

         if ( op ==  ) tree[rt].sumx = max( tree[rt].sumx, key );

         else tree[rt].sumy = max( tree[rt].sumy, key );

         return ;

     }

     int m = tree[rt].mid();

     if ( pos <= m ) update( op, pos, key, rtl );

     else update( op, pos, key, rtr );

     if ( op ==  ) tree[rt].sumx = max( tree[rtl].sumx, tree[rtr].sumx );

     else tree[rt].sumy = max( tree[rtl].sumy, tree[rtr].sumy );

 }

 int query( int op, int L, int R, int rt ) {

     if ( L <= tree[rt].l && tree[rt].r <= R ) {

         if ( op ==  ) return tree[rt].sumx;

         else return tree[rt].sumy;

     }

     int m = tree[rt].mid();

     if ( L > m ) return query( op, L, R, rtr );

     else if ( R <= m ) return query( op, L, R, rtl );

     else return max( query( op, L, m, rtl ), query( op, m + , R, rtr ) );

 }

 int L[maxn], R[maxn], n, sum[maxn];

 int main() {

     sf( n );

     int tot = ;

     for ( int i =  ; i <= n ; i++ ) {

         sff( L[i], R[i] );

         sum[tot++] = L[i], sum[tot++] = R[i];

     }

     sort( sum, sum + tot );

     tot = unique( sum, sum + tot ) - sum;

     build( , tot,  );

     LL ans = ;

     for ( int i = n ; i >=  ; i-- ) {

         int a = L[i], b = R[i];

         L[i] = lower_bound( sum, sum + tot, L[i] ) - sum + ;

         R[i] = lower_bound( sum, sum + tot, R[i] ) - sum + ;

         int ny = query( , L[i], tot,  );

         int nx = query( , R[i], tot,  );

         ans += b - ny;

         ans += a - nx;

      //   fuck(nx),fuck(ny);

         update( , L[i], sum[R[i] - ],  );

         update( , R[i], sum[L[i] - ],  );

     }

     printf( "%lld\n", ans );

     return ;

 }