hdu3275 线段树区间更新（略坑

先附上题目

Description

Felicia was given a long string of fancy lanterns, but not all of them are on to light the night. Felicia felt bad about that and he wants to light up all the lanterns. There is a kind of magic switch which can change the states of k continuous lanterns. Once you choose k continuous lanterns and install a switch on them, the states of all k continuous lanterns can be changed together (on ->off or off ->on), but you cannot choose some ones be changed and some ones not be changed. 
Felicia wants to buy as few switches as possible so that he can install them to turn on all the lanterns. Please notice that each switch must control exactly k continuous lanterns. 

Input

The input consists of multiple test cases. The first line of each case contains integers n(0 < n <= 100000), which is the length of that string of fancy lanterns, and k(0 <= k <= n), which is the number of continuous lanterns that a switch will control. The next line consists of a string of “01” with length n. “1” means that lantern is on and “0” means off. Your job is turn all the “0” to “1”. 
The last test case is followed by a line containing two zeros. 

Output

If you cannot finish this job, output “-1” or you should output the minimum number switches that you should buy to turn on all the lanterns.

Sample Input

4 200004 210014 201104 201110 0

Sample Output

213-1

题意很简单   给你一串长度为n只有0,1的字符串，每次操作固定操作m长度的子串   问至少多少次才能使字符串全为 1

一开始没啥思路   后来发现每次找到第一个  0  的位置再以他为起点进行更新，重复操作要么全为 1   要么无法更新（别问我为什么，这是我队友跟我说的，我负责码线段树

这题我觉得有两个坑点

1.当m==0 的时候   这还能玩？？一个字符都不让动    出题的人也是脑残，，，总是动一些小条件      总之这时候就输出他的其实状态 ，即如果全为 1  输出0  否则 输出-1

2.延迟更新的时候，这个就需要对线段树有一定的理解了    因为是延迟更新，在还未更新的时候可能存在对某一个区间多次标记    标记奇数次次就是0,1反转，偶数次就是不变咯   这里我对lazy标记进行异或，一切迎刃而解

ac code

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define root tree[id]#define lson tree[id<<1]#define rson tree[id<<1|1]using namespace std;const int maxn=100005;int st,en;struct segtree{    int le,ri;    int sum,len,d;} tree[maxn*4];char ch[maxn];void pushdown(int id){    if(root.d&&root.le!=root.ri)    {        lson.sum=lson.len-lson.sum;        rson.sum=rson.len-rson.sum;        lson.d^=1;        rson.d^=1;        root.d=0;    }}void build_tree(int id,int le,int ri){    root.le=le,root.ri=ri,root.len=ri-le+1,root.d=0;    if(le==ri)    {        root.sum=ch[le]-'0';        return ;    }    int mid=(le+ri)>>1;    build_tree(id<<1,le,mid);    build_tree(id<<1|1,mid+1,ri);    root.sum=lson.sum+rson.sum;}int query(int id,int le,int ri){    int ans;    if(root.len==root.sum) return 0;    if(le==ri) return le;    pushdown(id);    int mid=(le+ri)>>1;    if(ans=query(id<<1,le,mid)) return ans;    if(ans=query(id<<1|1,mid+1,ri)) return ans;    root.sum=lson.sum+rson.sum;    return 0;}void update(int id,int le,int ri){    if(st<=le&&ri<=en)    {        root.sum=root.len-root.sum;        root.d^=1;        return ;    }    pushdown(id);    int mid=(le+ri)>>1;    if(st<=mid) update(id<<1,le,mid);    if(en>mid) update(id<<1|1,mid+1,ri);    root.sum=lson.sum+rson.sum;}int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(m==0&&n==0) break;        getchar();        scanf("%s",ch+1);        build_tree(1,1,n);        if(m==0)        {            if(tree[1].sum==tree[1].len) printf("0\n");            else printf("-1\n");            continue;        }        bool flag=true;        int step=0;        while(st=query(1,1,n))        {            en=st+m-1;            if(en>n)            {                flag=false;                break;            }            step++;            update(1,1,n);        }        if(flag) printf("%d\n",step);        else printf("-1\n");    }    return 0;}