线段树（区间更新与区间查询）——Just a Hook ( HDU 1698 )

• 题目链接：
  http://acm.hdu.edu.cn/showproblem.php?pid=1698

• 分析&题解：不多说，这类水题都放模板

1.线段树标准预处理：

#define lc (d<<1)
#define rc (d<<1|1)
#define mid ( l+r >>1)
//l,r表示整个区间范围

2.建树：

#define Max 100005

int a[Max];

struct Tr
{
    int sum, lz;//区间和，懒操作
}tr[Max<<2];

void Push(int root)
{
    tr[d].sum = tr[lc].sum + tr[rc].sum;
}

void build (int root, int l, int r)
{
    tr[root].lz = 0;
    if( l == r )
    {
        tr[root].sum = a[l];
        return ;
    }
    build(lc, l, mid);
    build(rc, mid+1, r);
    Push(root);
}

2.懒操作：因为是区间求和，所以区间更新后需要一步懒操作，更新整个区间的和：

void lazy(int root, int l, int r)
{//因为是更新操作，所以直接赋值
    if(tr[root].lz)
    {
        tr[lc].lz = tr[root].lz;
        tr[rc].lz = tr[root].lz;
        tr[lc].sum = tr[root].lz*(mid-l+1);
        tr[rc].sum = tr[root].lz*(r-mid);
        tr[root].lz = 0;
    }
}

3.区间更新：

void update (int root,  int l,  int r,  int L,  int R,  int k)
{
    if(l == L && r == R)//因为是更新操作，所以直接赋值
    {
        tr[root].lz = k;
        tr[root].sum = k*(r-l+1);
        return ;
    }
    lazy(root, l, r);
    if( R <= mid) update(lc, l, mid, L, R, k);
    else if(L > mid) update( rc, mid+1, r, L, R, k);
    else
    {
        update(lc, l, mid, L, mid, k);
        update(rc, mid+1, r, mid+1, R, k);
    }
    Push(root);
}

4.区间查询：

int query(int root,  int l ,  int r,  int L,  int R )
{
    if( l == L && r == R)
    {
        return tr[root].sum;
    }
    lazy(d, l, r);
    if( R <= mid) return query(lc, l, mid, L, R);
    else if (L > mid) return query( rc, mid+1, r, L, R);
    else return query(lc, l, mid, L, mid) + query(rc, mid+1, r, mid+1, R);
}

• AC代码：

#include 
#include 
#include 
#include 

using namespace std;

#define lc (d<<1)
#define rc (d<<1|1)
#define mid ( l+r >>1)
#define Max 100005

int a[Max];

struct Tr
{
    int sum, lz;
}tr[Max<<2];

void Push(int d)
{
    tr[d].sum = tr[lc].sum + tr[rc].sum;
}

void build (int d, int l, int r)
{
    tr[d].lz = 0;
    if( l == r )
    {
        tr[d].sum = a[l];
        return ;
    }
    build(lc, l, mid);
    build(rc, mid+1, r);
    Push(d);
}

void lazy(int d, int l, int r)
{
    if(tr[d].lz)
    {
        tr[lc].lz = tr[d].lz;
        tr[rc].lz = tr[d].lz;
        tr[lc].sum = tr[d].lz*(mid-l+1);
        tr[rc].sum = tr[d].lz*(r-mid);
        tr[d].lz = 0;
    }
}

int query(int d,  int l ,  int r,  int L,  int R )
{
    if( l == L && r == R)
    {
        return tr[d].sum;
    }
    lazy(d, l, r);
    if( R <= mid) return query(lc, l, mid, L, R);
    else if (L > mid) return query( rc, mid+1, r, L, R);
    else return query(lc, l, mid, L, mid) + query(rc, mid+1, r, mid+1, R);
}

void update (int d,  int l,  int r,  int L,  int R,  int k)
{
    if(l == L && r == R)
    {
        tr[d].lz = k;
        tr[d].sum = k*(r-l+1);
        return ;
    }
    lazy(d, l, r);
    if( R <= mid) update(lc, l, mid, L, R, k);
    else if(L > mid) update( rc, mid+1, r, L, R, k);
    else
    {
        update(lc, l, mid, L, mid, k);
        update(rc, mid+1, r, mid+1, R, k);
    }
    Push(d);
}

int main()
{
    int T;
    scanf("%d", &T);
    int tt = 1;
    while(T--)
    {
        int N,Q;
        scanf("%d%d", &N, &Q);
        for(int i=1;i<=N;i++)
            a[i] = 1;
        build(1, 1, N);
        for(int i=1;i<=Q;i++)
        {
            int a,b,v;
            scanf("%d%d%d", &a,&b,&v);
            update(1, 1, N, a, b, v);
        }
        printf("Case %d: The total value of the hook is %d.\n", tt++, query(1, 1, N, 1, N));
    }


}