poj2892，线段树单点更新，区间合并

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

Hint

An illustration of the sample input:      OOOOOOOD 3   OOXOOOOD 6   OOXOOXOD 5   OOXOXXOR     OOXOOXOR     OOXOOOO

线段树做法:

#include <cstdio>#include <cstring>#include <stack>#include <iostream>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int maxn=50009;//lsum为以区间左端点为起始点的最长连续长度//rsum为以区间右端点为终点的最长连续长度//msum为整个区间的最长连续长度int lsum[maxn<<2],rsum[maxn<<2],msum[maxn<<2];int n,m;char op[10];void pushup(int rt,int l,int r){    lsum[rt]=lsum[rt<<1];    rsum[rt]=rsum[rt<<1|1];    msum[rt]=max(max(msum[rt<<1],msum[rt<<1|1]),rsum[rt<<1]+lsum[rt<<1|1]);    int m=l+r>>1;    //区间合并    if(msum[rt<<1]==m-l+1)        lsum[rt]+=lsum[rt<<1|1];    if(msum[rt<<1|1]==r-(m+1)+1)        rsum[rt]+=rsum[rt<<1];}void build(int l,int r,int rt){    lsum[rt]=rsum[rt]=msum[rt]=r-l+1;    if(l==r)        return;    int m=l+r>>1;    build(lson);    build(rson);}void update(int pos,int val,int l,int r,int rt){    if(l==r)    {        lsum[rt]=rsum[rt]=msum[rt]=val;        return;    }    int m=l+r>>1;    if(pos<=m)        update(pos,val,lson);    else update(pos,val,rson);    pushup(rt,l,r);}int query(int pos,int l,int r,int rt){    if(msum[rt]==r-l+1||msum[rt]==0||l==r)        return msum[rt];    int m=l+r>>1;    int ret;    if(pos<=m)    {       //如果查询点在左儿子区间，并且在左儿子区间的右端点最长连续范围内//那么就有可能存在右儿子区间中与其连续的段        if(pos>=m-rsum[rt<<1]+1)            ret=query(pos,lson)+query(m+1,rson);        else  ret=query(pos,lson);    }    else    {        if(pos<=m+1+lsum[rt<<1|1]-1)            ret=query(pos,rson)+query(m,lson);        else ret=query(pos,rson);    }    return ret;}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        build(1,n,1);        stack<int> st;        while(m--)        {            scanf("%s",op);            int tmp;            if(op[0]=='D')            {                scanf("%d",&tmp);                st.push(tmp);                update(tmp,0,1,n,1);            }            else if(op[0]=='Q')            {                scanf("%d",&tmp);                printf("%d\n",query(tmp,1,n,1));            }            else            {                tmp=st.top();                st.pop();                update(tmp,1,1,n,1);            }        }    }    return 0;}

stl做法：
对于要查询的点，用set找出其所在区间的上下界即可。

#include <cstdio>#include <set>#include <stack>using namespace std;int main(){    int n,m;    scanf("%d%d",&n,&m);   // getchar();    char ss[10];    set<int> s;    stack<int> st;    while(m--)    {        scanf("%s",ss);        int t;        if(ss[0]=='D')        {            scanf("%d",&t);            s.insert(t);            st.push(t);        }        else if(ss[0]=='Q')        {            scanf("%d",&t);            int l=1,r=n;            set<int>::iterator it;            if(s.find(t)!=s.end())                puts("0");            else            {                it=s.upper_bound(t);                if(it!=s.end())                {                    r=*(it)-1;                }                if(it!=s.begin())                {                    it--;                    l=*it+1;                }                printf("%d\n",r-l+1);            }        }        else        {            int tmp=st.top();            st.pop();            s.erase(tmp);        }    }    return 0;}