UVA 11992 - Fast Matrix Operations

题目链接：

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=501&page=show_problem&problem=3143

解题思路：

题目大意：给你一个矩阵，然后有3种操作：

1 x1 y1 x2 y2 val是将矩阵x1 y1 x2 y2的区间内的值全部增加val

2 x1 y1 x2 y2 val是将矩阵x1 y1 x2 y2的区间内的值全部设置为val

3 x1 y1 x2 y2 求出x1 y1 x2 y2的区间内的和，最大值，最小值。
矩阵不超过20行，矩阵元素却可能达到1000000个，可以想到每行建一颗线段树，则本题转为一维问题。
本题有两个操作，add和set，因此需要两个标记add和setv，含义同前。规定同时有两个标记时，表示先执行set再执行add，传递代码如下：

void pushdown(int id){
    if(tree[id].l >= tree[id].r)
        return;
    if(tree[id].setv){
        tree[id<<1].setv = tree[id<<1|1].setv = tree[id].setv;
        tree[id<<1].minn = tree[id<<1|1].minn = tree[id].setv;
        tree[id<<1].maxn = tree[id<<1|1].maxn = tree[id].setv;
        tree[id<<1].add = tree[id<<1|1].add = 0;
        tree[id<<1].sum = (tree[id<<1].r-tree[id<<1].l+1)*tree[id].setv;
        tree[id<<1|1].sum = (tree[id<<1|1].r-tree[id<<1|1].l+1)*tree[id].setv;
        tree[id].setv = 0;
    }
    if(tree[id].add){
        int tmp = tree[id].add;
        tree[id<<1].add += tmp;tree[id<<1|1].add += tmp;
        tree[id<<1].minn += tmp;tree[id<<1|1].minn += tmp;
        tree[id<<1].maxn +=tmp;tree[id<<1|1].maxn += tmp;
        tree[id<<1].sum += (tree[id<<1].r-tree[id<<1].l+1)*tmp;
        tree[id<<1|1].sum += (tree[id<<1|1].r - tree[id<<1|1].l+1)*tmp;
        tree[id].add = 0;
    }
}

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#define INF 0xfffffff
using namespace std;

const int N = 1000010;
struct node{
    int l,r;
    int minn,maxn,sum;
    int add,setv;
}tree[N<<2];

int anssum,ansmin,ansmax;

void maintain(int id){
    if(tree[id].l >= tree[id].r)
        return ;
    tree[id].maxn = max(tree[id<<1].maxn,tree[id<<1|1].maxn);
    tree[id].minn = min(tree[id<<1].minn,tree[id<<1|1].minn);
    tree[id].sum  = tree[id<<1].sum + tree[id<<1|1].sum;
}

void build(int id,int l,int r){//id:index
    tree[id].l = l;
    tree[id].r = r;
    tree[id].add = 0;
    tree[id].setv = 0;
    if(l == r){
        tree[id].sum = tree[id].minn = tree[id].maxn = 0;
        return;
    }
    int mid = (l+r)>>1;
    build(id<<1,l,mid);
    build(id<<1|1,mid+1,r);
    maintain(id);
}

void pushdown(int id){
    if(tree[id].l >= tree[id].r)
        return;
    if(tree[id].setv){
        tree[id<<1].setv = tree[id<<1|1].setv = tree[id].setv;
        tree[id<<1].minn = tree[id<<1|1].minn = tree[id].setv;
        tree[id<<1].maxn = tree[id<<1|1].maxn = tree[id].setv;
        tree[id<<1].add = tree[id<<1|1].add = 0;
        tree[id<<1].sum = (tree[id<<1].r-tree[id<<1].l+1)*tree[id].setv;
        tree[id<<1|1].sum = (tree[id<<1|1].r-tree[id<<1|1].l+1)*tree[id].setv;
        tree[id].setv = 0;
    }
    if(tree[id].add){
        int tmp = tree[id].add;
        tree[id<<1].add += tmp;tree[id<<1|1].add += tmp;
        tree[id<<1].minn += tmp;tree[id<<1|1].minn += tmp;
        tree[id<<1].maxn +=tmp;tree[id<<1|1].maxn += tmp;
        tree[id<<1].sum += (tree[id<<1].r-tree[id<<1].l+1)*tmp;
        tree[id<<1|1].sum += (tree[id<<1|1].r - tree[id<<1|1].l+1)*tmp;
        tree[id].add = 0;
    }
}

void updateSet(int id,int l,int r,int val){
    if(tree[id].l >= l && tree[id].r <= r){
        tree[id].setv = val;
        tree[id].minn = val;
        tree[id].maxn = val;
        tree[id].add = 0;
        tree[id].sum = (tree[id].r-tree[id].l+1)*val;
        return;
    }
    pushdown(id);
    int mid = (tree[id].l+tree[id].r)>>1;
    if(l <= mid)
        updateSet(id<<1,l,r,val);
    if(mid < r)
        updateSet((id<<1)+1,l,r,val);
    maintain(id);
}

void updateAdd(int id,int l,int r,int val){
    if(tree[id].l >= l && tree[id].r <= r){
        tree[id].add += val;
        tree[id].minn += val;
        tree[id].maxn += val;
        tree[id].sum += (tree[id].r-tree[id].l+1)*val;
        return;
    }
    pushdown(id);
    int mid = (tree[id].l+tree[id].r)>>1;
    if(l <= mid)
        updateAdd(id<<1,l,r,val);
    if(mid < r)
        updateAdd((id<<1)+1,l,r,val);
    maintain(id);
}

void query(int id,int l,int r){
     if(tree[id].l >= l && tree[id].r <= r){
        anssum += tree[id].sum;
        ansmin = min(ansmin,tree[id].minn);
        ansmax = max(ansmax,tree[id].maxn);
        return;
    }
    pushdown(id);
    int mid = (tree[id].l+tree[id].r)>>1;
    if(l <= mid)
        query(id<<1,l,r);
    if(mid < r)
        query(id<<1|1,l,r);
    maintain(id);
}

int main(){
    int r,c,m;
    while(~scanf("%d%d%d",&r,&c,&m)){
        build(1,1,r*c);
        int op,x1,x2,y1,y2,val;
        while(m--){
            scanf("%d",&op);
            if(op == 1){
                scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&val);
                for(int i = x1; i <= x2; i++)
                    updateAdd(1,(i-1)*c+y1,(i-1)*c+y2,val);
            }
            else if(op == 2){
                scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&val);
                for(int i = x1; i <= x2; i++)
                    updateSet(1,(i-1)*c+y1,(i-1)*c+y2,val);
            }
            else{
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                anssum = 0;ansmin = INF;ansmax = -INF;
                for(int i = x1; i <= x2; i++)
                    query(1,(i-1)*c+y1,(i-1)*c+y2);
                printf("%d %d %d\n",anssum,ansmin,ansmax);
            }
        }
    }
    return 0;
}

/*
4 4 8
1 1 2 4 4 5
3 2 1 4 4
1 1 1 3 4 2
3 1 2 4 4
3 1 1 3 4
2 2 1 4 4 2
3 1 2 4 4
1 1 1 4 3 3
*/