HDU1199 动态线段树 // 离散化

附动态线段树AC代码

http://acm.hdu.edu.cn/showproblem.php?pid=1199

由于昨天作了一道动态线段树的缘故，今天遇到了这题没有限制范围的题就天然而然想到了动态线段树的解法，写完看题解发现原来只要离散化就行了（干。。），总结了一下这题和昨天hdu5367的区别在于，虽然都是两题范围超级大的线段树，可是昨天的强制要求在线求解，只能选择空间复杂度更大一些的动态线段树来求解，而今天的这题能够选择离线操做，于是能够采用先读入全部输入，离散化以后建树的方法来操做。下次遇到这样的题仍是应当优先考虑离散化。

 

在一个所有涂黑色的条子上涂上一些白色或黑色的片断，问最大白色片断。

仅仅从线段树维护节点的角度上来看很简单，维护最大白色片断，左边最大白色片断，右边最大白色片断就行了。

因为条子的长度长达1-INT_MAX；

采用离散化，或者像我同样失了智去用动态线段树的方法，也能AC

#include <map> #include <set> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional>
#define For(i, x, y) for(int i=x; i<=y; i++)  
#define _For(i, x, y) for(int i=x; i>=y; i--)
#define Mem(f, x) memset(f, x, sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i = 0; i <= N ; i ++) u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
using namespace std; typedef vector<int> VI; const double eps = 1e-9; const int maxn = 2010; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; inline int read() { int now=0;register char c=getchar(); for(;!isdigit(c);c=getchar()); for(;isdigit(c);now=now*10+c-'0',c=getchar()); return now; } int N,M; struct Tree{ LL sum; //最大连续 
    LL lsum;   //左连续 
    LL rsum;   //右连续 
    int lt; int rt; int lazy; void init(){ lsum = rsum = sum = lt = rt = 0; lazy = -1; } }tree[maxn * 60]; int tot; void check(int &t){ if(t) return; t = ++tot; tree[t].init(); } void add(int &t,LL L,LL R,int v){ if(v){ tree[t].sum = tree[t].lsum = tree[t].rsum = R - L + 1; }else{ tree[t].sum = tree[t].lsum = tree[t].rsum = 0; } tree[t].lazy = v; } void Pushdown(int& t,LL L,LL R){ if(tree[t].lazy == -1) return; int &lt = tree[t].lt; int &rt = tree[t].rt; LL M = (L + R) >> 1; check(lt); check(rt); add(lt,L,M,tree[t].lazy); add(rt,M + 1,R,tree[t].lazy); tree[t].lazy = -1; } void Pushup(int t,LL L,LL R){ int &ls = tree[t].lt; int &rs = tree[t].rt; LL M = (L + R) >> 1; check(ls); check(rs); tree[t].sum = max(tree[ls].sum,tree[rs].sum); tree[t].sum = max(tree[t].sum,tree[ls].rsum + tree[rs].lsum); tree[t].lsum = tree[ls].lsum; if(tree[ls].lsum == M - L + 1){ tree[t].lsum = tree[ls].lsum + tree[rs].lsum; } tree[t].rsum = tree[rs].rsum; if(tree[rs].rsum == R - M){ tree[t].rsum = tree[rs].rsum + tree[ls].rsum; } } void update(int &t,int q1,int q2,LL L,LL R,int v){ check(t); if(q1 <= L && R <= q2){ add(t,L,R,v); return; } Pushdown(t,L,R); LL m = (L + R) >> 1; if(q1 <= m) update(tree[t].lt,q1,q2,L,m,v); if(q2 > m) update(tree[t].rt,q1,q2,m + 1,R,v); Pushup(t,L,R); } LL Left,Right; void query(int t,LL L,LL R){ if(L == R){ Left = L; Right = R; return; } check(tree[t].lt); check(tree[t].rt); int ls = tree[t].lt; int rs = tree[t].rt; LL M = (L + R) >> 1; if(tree[ls].sum == tree[t].sum) query(ls,L,M); else if(tree[rs].sum == tree[t].sum) query(rs,M + 1,R); else{ Left = M - tree[ls].rsum + 1; Right = M + tree[rs].lsum; return; } } int main() { while(~Sca(N)){ LL L = 1; LL R = 2147483647; tot = 0; int root = 0; update(root,L,R,L,R,0); For(i,1,N){ LL l,r; char op[3]; scanf("%lld%lld%s",&l,&r,&op); if(op[0] == 'w'){ update(root,l,r,L,R,1); }else{ update(root,l,r,L,R,0); } } if(!tree[root].sum){ puts("Oh, my god"); continue; } query(root,L,R); printf("%lld %lld\n",Left,Right); } return 0; }