UVA11992:Fast Matrix Operations(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;}