UVA11992:Fast Matrix Operations(线段树)

标签：uva   线段树   

题意：

有一个初始状态全为0的矩阵，一共有三个操作

1 x1 y1 x2 y2 v：子矩阵(x1,y1,x2,y2)所有元素增加v

2 x1 y1 x2 y2 v：子矩阵(x1,y1,x2,y2)所有元素设为v

3 x1 y1 x2 y2 v：查询子矩阵(x1,y1,x2,y2)的元素和，最大值和最小值

思路：

因为总元素葛素不超过10^6，而且更新是对于连续的行进行更新，所以我们可以把矩阵转化为一个一元组，通过下一行拼接在上一行的末尾，那么在更新与查询的时候只要对相应的区间进行操作即可

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define lson 2*i
#define rson 2*i+1
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 1000005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)

int ans_sum,ans_max,ans_min;
struct node
{
    int l,r;
    int sum,max,min;
    int add,set;
} a[N<<2];

void pushdown(int i)
{
    if(a[i].set!=-1)
    {
        a[lson].set = a[rson].set = a[i].set;
        a[lson].add = a[rson].add = 0;
        a[lson].min = a[rson].min = a[i].set;
        a[lson].max = a[rson].max = a[i].set;
        a[lson].sum = (a[lson].r-a[lson].l+1)*a[i].set;
        a[rson].sum = (a[rson].r-a[rson].l+1)*a[i].set;
        a[i].set = -1;
    }
    if(a[i].add>0)
    {
        a[lson].add+=a[i].add;
        a[rson].add+=a[i].add;
        a[lson].min+=a[i].add;
        a[rson].min+=a[i].add;
        a[lson].max+=a[i].add;
        a[rson].max+=a[i].add;
        a[lson].sum+=a[i].add*(a[lson].r-a[lson].l+1);
        a[rson].sum+=a[i].add*(a[rson].r-a[rson].l+1);
        a[i].add = 0;
    }
}

void pushup(int i)
{
    a[i].sum=a[lson].sum+a[rson].sum;
    a[i].max=max(a[lson].max,a[rson].max);
    a[i].min=min(a[lson].min,a[rson].min);
}

void build(int l,int r,int i)
{
    a[i].l = l;
    a[i].r = r;
    a[i].sum = 0;
    a[i].max = 0;
    a[i].min = 0;
    a[i].add = 0;
    a[i].set = -1;
    if(l == r) return;
    int mid = (l+r)>>1;
    build(LS);
    build(RS);
}

void set_data(int l,int r,int i,int val)
{
    if(a[i].l==l&&a[i].r==r)
    {
        a[i].sum = val*(r-l+1);
        a[i].min = val;
        a[i].max = val;
        a[i].set = val;
        a[i].add = 0;
        return;
    }
    pushdown(i);
    int mid = (a[i].l+a[i].r)>>1;
    if(r<=mid)
        set_data(l,r,lson,val);
    else if(l>mid)
        set_data(l,r,rson,val);
    else
    {
        set_data(LS,val);
        set_data(RS,val);
    }
    pushup(i);
}

void add_data(int l,int r,int i,int val)
{
    if(a[i].l==l&&a[i].r==r)
    {
        a[i].sum += val*(r-l+1);
        a[i].min += val;
        a[i].max += val;
        a[i].add += val;
        return;
    }
    pushdown(i);
    int mid = (a[i].l+a[i].r)>>1;
    if(r<=mid)
        add_data(l,r,lson,val);
    else if(l>mid)
        add_data(l,r,rson,val);
    else
    {
        add_data(LS,val);
        add_data(RS,val);
    }
    pushup(i);
}

void query(int l,int r,int i)
{
    if(l == a[i].l && a[i].r == r)
    {
        ans_sum += a[i].sum;
        ans_max = max(ans_max,a[i].max);
        ans_min = min(ans_min,a[i].min);
        return ;
    }
    pushdown(i);
    int mid = (a[i].l+a[i].r)>>1;
    if(r<=mid)
        query(l,r,lson);
    else if(l>mid)
        query(l,r,rson);
    else
    {
        query(LS);
        query(RS);
    }
    pushup(i);
}

int main()
{
    int op,i,j,x1,x2,y1,y2,v;
    int r,c,m;
    while(~scanf("%d%d%d",&r,&c,&m))
    {
        build(1,r*c,1);
        while(m--)
        {
            scanf("%d",&op);
            if(op==1)
            {
                scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&v);
                for(i = x1; i<=x2; i++)
                    add_data((i-1)*c+y1,(i-1)*c+y2,1,v);
            }
            else if(op==2)
            {
                scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&v);
                for(i = x1; i<=x2; i++)
                    set_data((i-1)*c+y1,(i-1)*c+y2,1,v);
            }
            else
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                ans_sum = 0;
                ans_max = -INF;
                ans_min = INF;
                for(i = x1; i<=x2; i++)
                    query((i-1)*c+y1,(i-1)*c+y2,1);
                printf("%d %d %d\n",ans_sum,ans_min,ans_max);
            }
        }
    }

    return 0;
}

UVA11992:Fast Matrix Operations(线段树)

标签：uva   线段树   

原文地址：http://blog.csdn.net/libin56842/article/details/46489841