• poj 3468 A Simple Problem with Integers


    #include<stdio.h>
    struct CNode
    {
        int L,R;
        long long nSum;
        long long Inc;
        CNode *pLeft,*pRight;
    };
    CNode Tree[1000000];
    int nCount;
    int Mid(CNode *pRoot)
    {
        return (pRoot->L+pRoot->R)/2;
    }
    void BuildTree(CNode *pRoot,int L,int R)
    {
        pRoot->L=L;
        pRoot->R=R;
        pRoot->nSum=0;
        pRoot->Inc=0;
        if(L==R)
            return ;
        nCount++;
        pRoot->pLeft=Tree+nCount;
        nCount++;
        pRoot->pRight=Tree+nCount;
        BuildTree(pRoot->pLeft,L,(L+R)/2);
        BuildTree(pRoot->pRight,(L+R)/2+1,R);
    }
    void Insert(CNode *pRoot,int i,int v)
    {
        if(pRoot->L==i&&pRoot->R==i)
        {
            pRoot->nSum=v;
            return ;
        }
        pRoot->nSum+=v;
        if(i<=Mid(pRoot))
            Insert(pRoot->pLeft,i,v);
        else
            Insert(pRoot->pRight,i,v);
    }
    void Add(CNode *pRoot,int a,int b,long long c)
    {
        if(pRoot->L==a&&pRoot->R==b)
        {
            pRoot->Inc+=c;
            return ;
        }
        pRoot->nSum+=c*(b-a+1);
        if(b<=Mid(pRoot))
            Add(pRoot->pLeft,a,b,c);
        else if(a>Mid(pRoot))
            Add(pRoot->pRight,a,b,c);
        else
        {
            Add(pRoot->pLeft,a,Mid(pRoot),c);
            Add(pRoot->pRight,Mid(pRoot)+1,b,c);
        }
    }
    long long QuerySum(CNode *pRoot,int a,int b)
    {
        if(pRoot->L==a&&pRoot->R==b)
        {
            return pRoot->nSum+(pRoot->R-pRoot->L+1)*pRoot->Inc;
        }
        pRoot->nSum+=(pRoot->R-pRoot->L+1)*pRoot->Inc;
        Add(pRoot->pLeft,pRoot->L,Mid(pRoot),pRoot->Inc);
        Add(pRoot->pRight,Mid(pRoot)+1,pRoot->R,pRoot->Inc);
        pRoot->Inc=0;
        if(b<=Mid(pRoot))
            return QuerySum(pRoot->pLeft,a,b);
        else if(a>Mid(pRoot))
            return QuerySum(pRoot->pRight,a,b);
        else
        {
            return QuerySum(pRoot->pLeft,a,Mid(pRoot))+
            QuerySum(pRoot->pRight,Mid(pRoot)+1,b);
        }
    }
    int main()
    {
        int n,q,a,b,c;
        int i,j,k;
        char cmd[10];
        scanf("%d%d",&n,&q);
        nCount=0;
        BuildTree(Tree,1,n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a);
            Insert(Tree,i,a);
        }
        for(i=0;i<q;i++)
        {
            scanf("%s",cmd);
            if(cmd[0]=='C')
            {
                scanf("%d%d%d",&a,&b,&c);
                Add(Tree,a,b,c);
            }
            else
            {
                scanf("%d%d",&a,&b);
                printf("%d
    ",QuerySum(Tree,a,b));
            }
        }
        return 0;
    }

    在增加时,如果要加的区间正好覆盖一个
    节点,则增加其节点的Inc值,不再往下走
    ,否则要更新nSum,再将增量往下传
    这样更新的复杂度就是O(log(n))
    在查询时,如果待查区间不是正好覆盖一
    个节点,就将节点的Inc往下带,然后将Inc
    代表的所有增量累加到nSum上后将Inc清0
    ,接下来再往下查询。 Inc往下带的过程也是
    区间分解的过程,复杂度也是O(log(n))

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  • 原文地址:https://www.cnblogs.com/XDJjy/p/3273492.html
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