DynamicSegmentTree

最近尝试了一下动态开点线段树，英文直译就是Dynamic Open Point Segment Tree，太SB了。

就跟之前的主席树写法差不多。

 if(!x || x == y) {
     x = ++tot;
 }

主席树

 if(!o) {
     o = ++tot;
 }

动态开点线段树

实际上当普通线段树用就行了......

例题：洛谷P1908 逆序对

我们可以不离散化，使用动态开点，直接在1e9上开值域线段树。吸氧后堪堪A掉。

 // luogu-judger-enable-o2
 #include <cstdio>

 typedef long long LL;
 const int N = ;

 int sum[N * ], tot, ls[N * ], rs[N * ], a[N], root;

 void add(int L, int R, int l, int r, int &o) {
     if(!o) {
         o = ++tot;
     }
     if(L <= l && r <= R) {
         sum[o]++;
         return;
     }
     int mid = (l + r) >> ;
     if(L <= mid) {
         add(L, R, l, mid, ls[o]);
     }
     if(mid < R) {
         add(L, R, mid + , r, rs[o]);
     }
     sum[o] = sum[ls[o]] + sum[rs[o]];
     return;
 }

 int ask(int L, int R, int l, int r, int o) {
     if(!o) {
         return ;
     }
     if(L <= l && r <= R) {
         return sum[o];
     }
     int mid = (l + r) >> ;
     int ans = ;
     if(L <= mid) {
         ans += ask(L, R, l, mid, ls[o]);
     }
     if(mid < R) {
         ans += ask(L, R, mid + , r, rs[o]);
     }
     return ans;
 }

 int main() {
     //printf("%d", sizeof(sum) / 1024 / 1024 * 3);
     int n, m = 1e9;

     scanf("%d", &n);
     for(int i = ; i <= n; i++) {
         scanf("%d", &a[i]);
     }

     LL ans = ;
     for(int i = ; i <= n; i++) {
         ans += ask(a[i] + , m, , m, );
         add(a[i], a[i], , m, root);
     }
     printf("%lld", ans);
     return ;
 }

AC代码

例题：洛谷P2471 降雨量

这个题的难点TM在于分类讨论......大毒瘤。

对于未知的位置，我们用一个bool数组维护就行了。

询问的时候查x，y，(x,y)三段，然后暴力讨论.....

 #include <cstdio>
 #include <algorithm>

 const int N = , E = 1e9 + ;

 int root, large[N * ], tot, ls[N * ], rs[N * ];
 bool maybe, sure[N * ];

 void change(int p, int v, int l, int r, int &o) {
     //printf("change : %d %d %d %d \n", p, v, l, r);
     //getchar();
     if(!o) {
         o = ++tot;
     }
     if(l == r) {
         large[o] = v;
         sure[o] = ;
         return;
     }
     int mid = l + (r - l) / ;
     if(p <= mid) {
         change(p, v, l, mid, ls[o]);
     }
     else {
         change(p, v, mid + , r, rs[o]);
     }
     large[o] = std::max(large[ls[o]], large[rs[o]]);
     sure[o] = sure[ls[o]] & sure[rs[o]];
     return;
 }

 int ask(int L, int R, int l, int r, int o) {
     //printf("%d %d %d %d \n", L, R, l, r);
     if(!o) {
         maybe = ;
         return ;
     }
     if(L <= l && r <= R) {
         if(!sure[o]) {
             maybe = ;
         }
         return large[o];
     }
     int mid = l + (r - l) / ;
     int ans = ;
     if(L <= mid) {
         ans = std::max(ans, ask(L, R, l, mid, ls[o]));
     }
     if(mid < R) {
         ans = std::max(ans, ask(L, R, mid + , r, rs[o]));
     }
     return ans;
 }

 int main() {
     int n, limit = E << , x, y;
     scanf("%d", &n);
     for(int i = ; i <= n; i++) {
         scanf("%d%d", &x, &y);
         change(x + E, y, , limit, root);
     }
     scanf("%d", &n);
     while(n--) {
         scanf("%d%d", &y, &x);
         if(y == x) {
             printf("true\n");
             continue;
         }
         int c = ask(y + E, y + E, , limit, root);
         int cc = maybe;
         maybe = ;
         int d = ask(x + E, x + E, , limit, root);
         int dd = maybe;
         maybe = ;
         //printf("%d %d %d %d \n", c, cc, d, dd);    false -> maybe
         if(y +  == x) {
             if(cc || dd){
                 printf("maybe\n");
             }
             else {
                 if(c >= d) {
                     printf("true\n");
                 }
                 else {
                     printf("false\n");
                 }
             }
             continue;
         }
         int e = ask(y +  + E, x -  + E, , limit, root);
         int ee = maybe;
         maybe = ;
         //printf("%d %d \n", e, ee);
         if(dd) {
             if(cc) {
                 printf("maybe\n");
             }
             else {
                 if(e >= c) {
                     printf("false\n");
                 }
                 else {
                     printf("maybe\n");
                 }
             }
             continue;
         }
         if(cc) {
             if(e >= d) {
                 printf("false\n");
             }
             else {
                 printf("maybe\n");
             }
             continue;
         }
         if(ee) {
             if(e >= d || c < d) {
                 printf("false\n");
             }
             else {
                 printf("maybe\n");
             }
             continue;
         }
         if(e >= d || c < d) {
             printf("false\n");
         }
         else {
             printf("true\n");
         }
     }

     return ;
 }

AC代码