【线段树区间修改】fzu2105Digits Count

标签：os   io   ar   amp   type   ios   ef   har   

/*
题意：
给出数组A，有以下几个操作：
1: AND(opn, L, R):把区间[L, R]中的元素A[i]改为A[i] & opn;;;;;;
2: OR(opn, L, R) :把区间[L, R]中的元素A[i]改为A[i] | opn;;;;;;;
3: XOR(opn, L, R):把区间[L, R]中的元素A[i]改为A[i] ^ opn;;;;;;;
4: SUM(L, R)     :对区间[L, R]中的元素求和；；；；
--------------------------------------------------------------------------------
线段树区间修改的题目：
--------------------------------------------------------------------------------
void build(int l, int r, int o)递归建树
{
    tree[o].l = l;结点的左端点
    tree[o].r = r;结点的右端点
    tree[o].num = -1;标记该区间的数是否相同
    if(l == r)叶子结点
    {
        scanf("%d",&tree[o].num);
        return ;
    }
    int M = (l + r)/2;
    build(l, M, lc);
    build(M+1, r, rc);
    if(tree[lc].num != -1 && tree[lc].num == tree[rc].num)向上更新
        tree[o].num = tree[lc].num;
}
------------------------------------------------------------------------------------
1,2,3操作时数组的更新：
int opreate(int op, int opn, int num)对应的操作
{
    if(op == 1) return num & opn;------------AND
    if(op == 2) return num | opn;------------OR
    if(op == 3) return num ^ opn;------------XOR
}

void update(int l, int r, int o, int opn, int op)
{
    if(tree[o].l == l && tree[o].r == r && tree[o].num >= 0)找到该区间时，跟新区间内的元素，，即修改标记
    {
        tree[o].num = opreate(op, opn, tree[o].num);
        return ;
    }
    if(tree[o].num >= 0)标记传递，起作用是把num的值向下传递即：pushdown函数
    {
        tree[lc].num = tree[rc].num = tree[o].num;
        tree[o].num = -1;清除本节点的标记
    }
    int mid = (tree[o].l + tree[o].r)/2;
    if(r <= mid)------------------------------------------------------------？
        update(l, r, lc, opn, op);更新左子树
    else if(l > mid)
        update(l, r, rc, opn, op);更新右子树
    else否则都要更新
    {
        update(l, mid, lc, opn, op);
        update(mid+1, r, rc, opn, op);
    }------------------------------------------------------------------------？
    if(tree[lc].num != -1 && tree[lc].num == tree[rc].num)往上更新
        tree[o].num = tree[lc].num;
}
----------------------------------------------------------------------------------------
LL query(int l, int r, int o)
{
    if(tree[o].l == l && tree[o].r == r && tree[o].num >= 0)
        return tree[o].num*(tree[o].r - tree[o].l+1);
    if(tree[o].num >= 0)
    {
        tree[lc].num = tree[rc].num = tree[o].num;
        tree[o].num = -1;
    }
    int mid = (tree[o].l + tree[o].r)/2;
    if(r <= mid)
        return query(l, r, lc);
    else if( l > mid)
        return query(l, r, rc);
    else
        return query(l, mid, lc)+query(mid+1, r, rc);
     if(tree[lc].num != -1 && tree[lc].num == tree[rc].num)
        tree[o].num = tree[lc].num;
}
--------------------------------------------------------------------------------------------
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f
#define lc o*2
#define rc o*2+1

using namespace std;

const int MAXN = 1000010;
typedef long long LL;

int n, m;
int a[MAXN];

struct node{
    int l, r;
    int num;
}tree[MAXN*4];

void build(int l, int r, int o)
{
    tree[o].l = l;
    tree[o].r = r;
    tree[o].num = -1;
    if(l == r)
    {
        scanf("%d",&tree[o].num);
        return ;
    }
    int M = (l + r)/2;
    build(l, M, lc);
    build(M+1, r, rc);
    if(tree[lc].num != -1 && tree[lc].num == tree[rc].num)
        tree[o].num = tree[lc].num;
}

int opreate(int op, int opn, int num)
{
    if(op == 1) return num & opn;
    if(op == 2) return num | opn;
    if(op == 3) return num ^ opn;
}

void update(int l, int r, int o, int opn, int op)
{
    if(tree[o].l == l && tree[o].r == r && tree[o].num >= 0)
    {
        tree[o].num = opreate(op, opn, tree[o].num);
        return ;
    }
    if(tree[o].num >= 0)
    {
        tree[lc].num = tree[rc].num = tree[o].num;
        tree[o].num = -1;
    }
    int mid = (tree[o].l + tree[o].r)/2;
    if(r <= mid)
        update(l, r, lc, opn, op);
    else if(l > mid)
        update(l, r, rc, opn, op);
    else
    {
        update(l, mid, lc, opn, op);
        update(mid+1, r, rc, opn, op);
    }
    if(tree[lc].num != -1 && tree[lc].num == tree[rc].num)
        tree[o].num = tree[lc].num;
}

LL query(int l, int r, int o)
{
    if(tree[o].l == l && tree[o].r == r && tree[o].num >= 0)
        return tree[o].num*(tree[o].r - tree[o].l+1);
    if(tree[o].num >= 0)
    {
        tree[lc].num = tree[rc].num = tree[o].num;
        tree[o].num = -1;
    }
    int mid = (tree[o].l + tree[o].r)/2;
    if(r <= mid)
        return query(l, r, lc);
    else if( l > mid)
        return query(l, r, rc);
    else
        return query(l, mid, lc)+query(mid+1, r, rc);
     if(tree[lc].num != -1 && tree[lc].num == tree[rc].num)
        tree[o].num = tree[lc].num;
}

int main()
{
    freopen("input.txt","r",stdin);
    char op[5];
    int opn, L, R;
    int T;
    cin>>T;
    while(T--)
    {
        scanf("%d%d", &n,&m);
        build(1, n, 1);
        while(m--)
        {
            scanf("%s",op);
            getchar();
            if(op[0] == 'S')
            {
                scanf("%d%d",&L, &R);
                printf("%lld\n",query(L+1, R+1, 1));
            }
            else
            {
                scanf("%d%d%d",&opn, &L, &R);
                if(op[0] == 'A')
                    update(L+1, R+1, 1, opn, 1);
                if(op[0] == 'O')
                    update(L+1, R+1, 1, opn, 2);
                if(op[0] == 'X')
                    update(L+1, R+1, 1, opn, 3);
            }
        }
    }
    return 0;
}

---------------------------------------------------------------------

wuwuwuwuuuuwuuwuwuuwuwuwuwuuw....................................

【线段树区间修改】fzu2105Digits Count,布布扣,bubuko.com

【线段树区间修改】fzu2105Digits Count

标签：os   io   ar   amp   type   ios   ef   har   

原文地址：http://blog.csdn.net/u013147615/article/details/38444603