poj-1540---Tunnel Warfare （线段树）

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9004    Accepted Submission(s): 3474

Problem Description

During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village was directly connected with two neighboring ones.

Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately!

 

Input

The first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event.

There are three different events described in different format shown below:

D x: The x-th village was destroyed.

Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.

R: The village destroyed last was rebuilt.

 

Output

Output the answer to each of the Army commanders’ request in order on a separate line.

 

Sample Input

7 9D 3D 6D 5Q 4Q 5RQ 4RQ 4

 

Sample Output

1024

 

Source

POJ Monthly

 

这题就是一个线段树的单点更新然后涉及到查询最大连续区间。

题意：输入n，m，表示有n个点，m次操作，D表示摧毁当前点，Q表示查询该点及该点左右两边未被摧毁的最大连续区间长度，R表示将最近一次被摧毁的点复原。

思路：拿到题第一想法便是线段树，但是关于线段树维护的是什么想了好久。这题最后我写的线段树维护了三个值，sum，lsum，rsum，分别表示当前区间最大的连续长度，当前区间从左开始的最大连续长度和从右开始的最大连续长度，如过在查询时查询点在这三个的区间内便可以直接返回了（想想看是不是这样，代码中解释了），然后涉及到的就是线段树的单点更新，比较简单。这道题总体来说不算难，跟写过的Hotile很像，据说暴力也可以过！但是暴力姿势得正确，我看了暴力代码。基本是卡着时间过的。

ac代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stack>using namespace std;int n,m;struct node{    int l,r;    int lsum,rsum,sum;//当前区间最大左连续值，最大右连续值，最大连续值}segtree[200005];stack<int> s;void pushup(int root){    segtree[root].lsum=segtree[root<<1].lsum;    segtree[root].rsum=segtree[root<<1|1].rsum;    segtree[root].sum=max(max(segtree[root<<1].sum,segtree[root<<1|1].sum),segtree[root<<1].rsum+segtree[root<<1|1].lsum);    if(segtree[root<<1].lsum==segtree[root<<1].r-segtree[root<<1].l+1)//如果左儿子的lsun等于区间长度        segtree[root].lsum=segtree[root<<1].sum+segtree[root<<1|1].lsum;//则当前区间的lsum等于左儿子的sum+右儿子的laum    if(segtree[root<<1|1].rsum==segtree[root<<1|1].r-segtree[root<<1|1].l+1)//同上        segtree[root].rsum=segtree[root<<1|1].sum+segtree[root<<1].rsum;}void buildtree(int root,int l,int r){    segtree[root].l=l;    segtree[root].r=r;    if(l==r)    {        segtree[root].sum=segtree[root].lsum=segtree[root].rsum=1;//初始值应该都为1        return ;    }    int mid=(l+r)>>1;    buildtree(root<<1,l,mid);    buildtree(root<<1|1,mid+1,r);    pushup(root);}void update(int root,int id,int tag)//tag为摧毁或者复原的标记{    int ll=segtree[root].l;    int rr=segtree[root].r;    if(ll==rr)    {        if(tag==1)//摧毁该点            segtree[root].sum=segtree[root].lsum=segtree[root].rsum=0;        if(tag==0)//修复该点            segtree[root].sum=segtree[root].lsum=segtree[root].rsum=1;        return;    }    int mid=(ll+rr)>>1;    if(id<=mid)update(root<<1,id,tag);    else update(root<<1|1,id,tag);    pushup(root);}int query(int root,int id){    int ll=segtree[root].l;    int rr=segtree[root].r;    if(ll==rr)        return segtree[root].sum;    if(segtree[root].sum==segtree[root].r-segtree[root].l+1)//当前区间sum值为区间长度则可以直接返回        return segtree[root].sum;    int mid=(ll+rr)>>1;    int ans=0;    if(id<=mid)    {        if(id>=segtree[root<<1].r-segtree[root<<1].rsum+1)//查询的点在左儿子的右连续区间里面，则返回左儿子的rsum和右儿子的lsum            return segtree[root<<1].rsum+segtree[root<<1|1].lsum;        else//否则在左儿子继续查询             return query(root<<1,id);    }    else    {        if(id<=segtree[root<<1|1].l+segtree[root<<1|1].lsum-1)//同上            return segtree[root<<1|1].lsum+segtree[root<<1].rsum;        else            return query(root<<1|1,id);    }}int main(){    while(~scanf("%d%d",&n,&m))    {        buildtree(1,1,n);        char str[3];        for(int i=1;i<=m;i++)        {            int x;            scanf("%s%d",str,&x);            if(str[0]=='D')            {                update(1,x,1);                s.push(x);//将摧毁的点入栈            }            else if(str[0]=='Q')            {                int ans=query(1,x);                cout<<ans<<endl;            }            else if(str[0]=='R')            {                int x=s.top();//取出最近摧毁的点                s.pop();                update(1,x,0);            }        }    }    return 0;}