ccnu-线段树-简单的区间更新（三题）

题目一：http://poj.org/problem?id=3468

Description

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.



Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.

The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.

Each of the next Q lines represents an operation.

"C abc" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.

"Q ab" means querying the sum of Aa, Aa+1, ... , Ab.



Output

You need to answer all Q commands in order. One answer in a line.



Sample Input

10 5

1 2 3 4 5 6 7 8 9 10

Q 4 4

Q 1 10

Q 2 4

C 3 6 3

Q 2 4

Sample Output

4

55

9

15

Hint

The sums may exceed the range of 32-bit integers.

裸的区间更新，关于lazy标记，还是老话，在下一次更新或询问时才把区间信息传递（pushdown）下去。

代码：

#include<iostream>
#include<cstdio>
#include<string>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Max  100020
long long a,b,q,n;
long long sum[Max<<2],c,lazy[Max<<2];
void pushup(int rt)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void pushdown(int rt,int m)
{
    if(lazy[rt])
    {
        lazy[rt<<1] += lazy[rt];
        lazy[rt<<1|1] += lazy[rt];
        sum[rt<<1] += lazy[rt] * (m - (m>>1));
        sum[rt<<1|1] += lazy[rt] * (m>>1);
        lazy[rt] = 0;
    }
}
void build(int l,int r,int rt)
{
    sum[rt] = 0;
    if(l == r)
    {
        scanf("%lld",&sum[rt]);
        return;
    }
    int m = (r + l)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
void Add(int L,int R,int v,int l,int r,int rt)
{
    if(L <= l&&r <= R)
    {
        lazy[rt] += v;
        sum[rt] += (long long)(r - l + 1) * v;
        return ;
    }
    pushdown(rt,r-l+1);
    int m = (r + l)>>1;
    if(L <= m) Add(L,R,v,lson);
    if(m < R) Add(L,R,v,rson);
    pushup(rt);
}
long long query(int L,int R,int l,int r,int rt)
{
    if(L <= l&&r <= R)
    {
        return sum[rt];
    }
    long long rec = 0;
    int m = (r + l)>>1;
    pushdown(rt,r-l+1);
    if(L <= m) rec += query(L,R,lson);
    if(m < R) rec += query(L,R,rson);
    return rec;
}
int main()
{
    scanf("%d%d",&n,&q);
    build(1,n,1);
    while(q--)
    {
        char qus[2];
        scanf("%s",&qus);
        if(qus[0] == 'C')
        {
            scanf("%d%d%d",&a,&b,&c);
            Add(a,b,c,1,n,1);
        }
        else
        {
            scanf("%d%d",&a,&b);
            printf("%lld\n",query(a,b,1,n,1));
        }

    }
    return 0;
}

题型二：http://acm.hdu.edu.cn/showproblem.php?pid=1698

把上题的求和改成更新区间值就可以了。

代码如下：

#include<iostream>
#include<cstdio>
#define Max 100010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int sum[Max<<2],lazy[Max<<2];
void pushup(int rt)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void pushdown(int rt,int INL)
{
    if(lazy[rt])
    {
        lazy[rt<<1] = lazy[rt<<1|1] = lazy[rt];
        sum[rt<<1] = lazy[rt] * (INL - (INL>>1));
        sum[rt<<1|1] = lazy[rt] * (INL>>1);
        lazy[rt] = 0;
    }
}
void build(int l,int r,int rt)
{
    lazy[rt] = 0;
    sum[rt] = 1;
    if(r == l)
    {
        return;
    }
    int m = (r + l) >>  1;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int L,int R,int val,int l,int r,int rt)
{
    if(L <= l&&r <= R)
    {
        sum[rt] = val * (r-l+1);
        lazy[rt] = val;
        return;
    }
    pushdown(rt,r-l+1);
    int m = (r + l)>>1;
    if(L <= m) update(L,R,val,lson);
    if(m < R) update(L,R,val,rson);
    pushup(rt);
}
int main()
{
    int T,n,x,y,z,q;
    scanf("%d",&T);
    int lala = T;
    while(T--)
    {
        scanf("%d",&n);
        build(1,n,1);
        scanf("%d",&q);
        while(q--)
        {
            scanf("%d%d%d",&x,&y,&z);
            update(x,y,z,1,n,1);
        }
        printf("Case %d: The total value of the hook is %d.\n",(lala-T),sum[1]);
    }
    return 0;
}

【题型三-hdu-ColorTheBall】http://acm.hdu.edu.cn/showproblem.php?pid=1556

题意：把a,b区间的 气球都涂上颜色，输出每个气球被涂色的次数。

思路：每次更新只记录到所更新的区间便不再往下记录更新，不需要pushup，只需要在输出时pushdown更新一次到每个节点就可以了

经验教训： 不要用cout， 不要用cout， 不要用cout！

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn 100010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int lazy[maxn<<4];
 int n;
//int index[maxn];
void build(int l, int r, int rt)
{
  int m;
  m = (l + r)>>1;
  lazy[rt] = 0;
  if(l == r) return;
  build(lson);
  build(rson);
}

void pushdown(int rt)
{
  if(lazy[rt])
  {
    lazy[rt<<1] += lazy[rt];
    lazy[rt<<1|1] += lazy[rt];
    lazy[rt] = 0;
  }
}

void paint(int l, int r, int rt, int L, int R)
{
  if(L <= l && r <= R)
  {
    lazy[rt]++;
    return ;
  }
  int m = (r+l)>>1;
  if(L <= m) paint(lson, L,R);
  if(m < R) paint(rson, L, R);
}

void finalQuery(int l, int r, int rt)
{
  int m = (l+r)>>1;
  if(l == r)
  {
     printf("%d%c",lazy[rt],l==n?'\n':' ');
     return;
  }
  pushdown(rt);   //在查询的时候才pushdown
  finalQuery(lson);
  finalQuery(rson);
}

int main()
{
 while(scanf("%d", &n) != EOF&&n)
 {
   build(1,n,1);
   for(int i = 0; i < n; i++)
   {
     int a, b;
     scanf("%d%d", &a, &b);
     paint(1, n, 1, a, b);
     //cout<<"-"<<endl;
   }
   finalQuery(1, n, 1);
 }
 return 0;
}