Tunnel Warfare（HDU1540+线段树+区间合并）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1540

题目：

题意：总共有n个村庄，有q次操作，每次操作分为摧毁一座村庄，修复一座村庄，和查询与询问的村庄相连接的城市数量。

思路：线段树+区间合并。

代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef unsigned long long ull;

 #define lson i<<1,l,mid
 #define rson i<<1|1,mid+1,r
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = 1e8;
 const int maxn = 5e4 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f3f;

 inline int read() {//读入挂
     int ret = , c, f = ;
     for(c = getchar(); !(isdigit(c) || c == '-'); c = getchar());
     if(c == '-') f = -, c = getchar();
     for(; isdigit(c); c = getchar()) ret = ret *  + c - '';
     if(f < ) ret = -ret;
     return ret;
 }

 int n, q, x, t;
 char op[];
 int tack[maxn];

 struct  node {
     int l, r, mx, ls, rs; //ls为左端最大连续区间，rs为右端最大连续区间，mx为区间内最大连续区间
 }segtree[maxn*];

 void build(int i, int l, int r) {
     segtree[i].l = l, segtree[i].r = r;
     segtree[i].mx = segtree[i].ls = segtree[i].rs = r - l + ;
     if(l == r) return;
     int mid = (l + r) >> ;
     build(lson);
     build(rson);
 }

 void update(int i, int pos, int x) {
     if(segtree[i].l == segtree[i].r) {
         if(x == ) segtree[i].mx = segtree[i].ls = segtree[i].rs = ; //此处为修复操作
         else segtree[i].mx = segtree[i].ls = segtree[i].rs = ; //摧毁操作
         return;
     }
     int mid = (segtree[i].l + segtree[i].r) >> ;
     if(pos <= mid) {
         update(i * , pos, x);
     } else {
         update(i *  + , pos, x);
     }
     segtree[i].ls = segtree[i*].ls;
     segtree[i].rs = segtree[i*+].rs;
     segtree[i].mx = max(max(segtree[i*].mx, segtree[i*+].mx), segtree[*i].rs + segtree[i*+].ls);
     if(segtree[i*].ls == segtree[i*].r - segtree[i*].l + ) { //如果左端全是连接的，那么ls还要右子树的ls
         segtree[i].ls += segtree[i*+].ls;
     }
     if(segtree[i*+].rs == segtree[i*+].r - segtree[i*+].l + ) { //同理
         segtree[i].rs += segtree[i*].rs;
     }
 }

 int query(int i, int pos) {
     if(segtree[i].l == segtree[i].r || segtree[i].mx == segtree[i].r - segtree[i].l +  || segtree[i].mx == ) {
         return segtree[i].mx;
     }
     int mid = (segtree[i].l + segtree[i].r) >> ;
     if(pos <= mid) {
         if(pos >= segtree[*i].r - segtree[i*].rs + ) { //如果pos与右边连接，那么还需要跑另一棵子树
             return query(i * , pos) + query(i *  + , mid + );
         } else {
             return query(i * , pos);
         }
     } else {
         if(pos <= segtree[i *  + ].l + segtree[i*+].ls - ) { //同理
             return query(i *  + , pos) + query(i * , mid);
         } else {
             return query(i *  + , pos);
         }
     }
 }

 int main() {
     //FIN;
     while(~scanf("%d%d", &n, &q)) {
         build(, , n);
         t = ;
         while(q--) {
             scanf("%s", op);
             if(op[] == 'D') {
                 scanf("%d", &x);
                 tack[++t] = x;
                 update(, x, );
             } else if(op[] == 'R') {
                 if(t > ) {
                     x = tack[t--];
                     update(, x, );
                 }
             } else {
                 scanf("%d", &x);
                 printf("%d\n", query(, x));
             }
         }
     }
     return ;
 }