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
2×1 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;}
在向山靠近一点，才发现这座山，好象一位诗人遥望远方，
