HDU 1540 Tunnel Warfare (线段树或set水过)

题意：D代表破坏村庄，R代表修复最后被破坏的那个村庄，Q代表询问包括x在内的最大连续区间是多少。

析：首先可以用set水过，set用来记录每个被破坏的村庄，然后查找时，只要查找左右两个端点好。

用线段树的话，就维护三个值分别是左端点连续右端点连续，全连续的最长的区别，然后用线段树维护就好。

代码如下：

set过：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

set<int> destroy;
stack<int> st;

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    destroy.clear();
    while(!st.empty())  st.pop();
    destroy.insert(0);
    destroy.insert(n+1);
    while(m--){
      char s[5];
      int x;
      scanf("%s", s);
      if(s[0] == ‘D‘){
        scanf("%d", &x);
        st.push(x);
        destroy.insert(x);
      }
      else if(s[0] == ‘R‘){
        if(st.empty())  continue;
        destroy.erase(st.top());
        st.pop();
      }
      else{
        scanf("%d", &x);
        if(destroy.count(x)){
          printf("0\n");  continue;
        }
        set<int> :: iterator it1 = destroy.lower_bound(x);
        set<int> :: iterator it2 = it1--;
        printf("%d\n", *it2-*it1-1);
      }
    }
  }
  return 0;
}

线段树：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

//0 - left 1 - right 2 - all
int sum[maxn<<2][3];

void push_up(int rt, int len){
  int l = rt<<1, r = rt<<1|1;
  sum[rt][0] = sum[l][0];
  sum[rt][1] = sum[r][1];
  sum[rt][2] = max(max(sum[l][2], sum[r][2]), sum[l][1]+sum[r][0]);
  if(sum[l][0] == len-len/2)  sum[rt][0] += sum[r][0];
  if(sum[r][1] == len/2)  sum[rt][1] += sum[l][1];
}

void build(int l, int r, int rt){
  sum[rt][0] = sum[rt][1] = sum[rt][2] = r - l + 1;
  if(l == r)  return ;
  int m = l + r >> 1;
  build(lson);
  build(rson);
}

void update(int M, int val, int l, int r, int rt){
  if(l == r){
    sum[rt][0] = sum[rt][1] = sum[rt][2] = val;
    return ;
  }
  int m = l + r >> 1;
  if(M <= m)  update(M, val, lson);
  else  update(M, val, rson);
  push_up(rt, r-l+1);
}

int query(int M, int l, int r, int rt){
  if(l == r || sum[rt][2] == 0 || sum[rt][2] == r-l+1)  return sum[rt][2];
  int m = l + r >> 1;
  if(M <= m){
    if(M > m - sum[rt<<1][1])  return query(M, lson) + sum[rt<<1|1][0];
    return query(M, lson);
  }
  if(M < m+1 + sum[rt<<1|1][0])  return query(M, rson) + sum[rt<<1][1];
  return query(M, rson);
}

stack<int> st;

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    build(1, n, 1);
    char s[10];
    while(!st.empty())  st.pop();
    while(m--){
      scanf("%s", s);
      int x;
      if(s[0] == ‘D‘){
        scanf("%d", &x);
        st.push(x);
        update(x, 0, 1, n, 1);
      }
      else if(s[0] == ‘R‘){
        if(st.empty())  continue;
        update(st.top(), 1, 1, n, 1);
        st.pop();
      }
      else{
        scanf("%d", &x);
        printf("%d\n", query(x, 1, n, 1));
      }
    }
  }
  return 0;
}