• UVA 11992


    UVA 11992 - Fast Matrix Operations

    题目链接

    题意:给定一个矩阵,3种操作,在一个矩阵中加入值a,设置值a。查询和

    思路:因为最多20列,所以全然能够当作20个线段树来做,然后线段树是区间改动区间查询,利用延迟操作,开两个延迟值一个存放set操作。一个存放add操作

    代码:

    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    #define lson(x) ((x<<1) + 1)
    #define rson(x) ((x<<1) + 2)
    #define INF 0x3f3f3f3f
    
    const int N = 1000005;
    
    int r, c, m;
    
    struct Node {
        int l, r;
        int sum, Max, Min, sumv, setv;
    } node[4 * N];
    
    void pushup(int x) {
        node[x].sum = node[lson(x)].sum + node[rson(x)].sum;
        node[x].Max = max(node[lson(x)].Max, node[rson(x)].Max);
        node[x].Min = min(node[lson(x)].Min, node[rson(x)].Min);
    }
    
    void pushdown(int x) {
        if (node[x].setv) {
    	node[lson(x)].sumv = node[rson(x)].sumv = 0;
    	node[lson(x)].setv = node[rson(x)].setv = node[x].setv;
    	node[lson(x)].sum = (node[lson(x)].r - node[lson(x)].l + 1) * node[x].setv;
    	node[rson(x)].sum = (node[rson(x)].r - node[rson(x)].l + 1) * node[x].setv;
    	node[lson(x)].Max = node[lson(x)].Min = node[x].setv;
    	node[rson(x)].Max = node[rson(x)].Min = node[x].setv;
    	node[x].setv = 0;
        }
        if (node[x].sumv) {
    	node[lson(x)].sumv += node[x].sumv;
    	node[rson(x)].sumv += node[x].sumv;
    	node[lson(x)].sum += (node[lson(x)].r - node[lson(x)].l + 1) * node[x].sumv;
    	node[rson(x)].sum += (node[rson(x)].r - node[rson(x)].l + 1) * node[x].sumv;
    	node[lson(x)].Max += node[x].sumv;
    	node[lson(x)].Min += node[x].sumv;
    	node[rson(x)].Max += node[x].sumv;
    	node[rson(x)].Min += node[x].sumv;
    	node[x].sumv = 0;
        }
    }
    
    void build(int l, int r, int x) {
        node[x].l = l; node[x].r = r;
        if (l == r) {
    	node[x].sum = node[x].Max = node[x].Min = node[x].sumv = node[x].setv = 0;
    	return;
        }
        int mid = (l + r) / 2;
        build(l, mid, lson(x));
        build(mid + 1, r, rson(x));
        pushup(x);
    }
    
    void add(int l, int r, int v, int x) {
        if (node[x].l >= l && node[x].r <= r) {
    	node[x].sumv += v;
    	node[x].sum += (node[x].r - node[x].l + 1) * v;
    	node[x].Max += v;
    	node[x].Min += v;
    	return;
        }
        pushdown(x);
        int mid = (node[x].l + node[x].r) / 2;
        if (l <= mid) add(l, r, v, lson(x));
        if (r > mid) add(l, r, v, rson(x));
        pushup(x);
    }
    
    void set(int l, int r, int v, int x) {
        if (node[x].l >= l && node[x].r <= r) {
    	node[x].setv = v;
    	node[x].sum = (node[x].r - node[x].l + 1) * v;
    	node[x].Max = node[x].Min = v;
    	node[x].sumv = 0;
    	return;
        }
        pushdown(x);
        int mid = (node[x].l + node[x].r) / 2;
        if (l <= mid) set(l, r, v, lson(x));
        if (r > mid) set(l, r, v, rson(x));
        pushup(x);
    }
    
    Node query(int l, int r, int x) {
        Node ans; ans.sum = 0; ans.Max = 0; ans.Min = INF;
        if (node[x].l >= l && node[x].r <= r) {
    	ans.sum = node[x].sum;
    	ans.Max = node[x].Max;
    	ans.Min = node[x].Min;
    	return ans;
        }
        pushdown(x);
        int mid = (node[x].l + node[x].r) / 2;
        if (l <= mid) {
    	Node tmp = query(l, r, lson(x));
    	ans.sum += tmp.sum;
    	ans.Max = max(ans.Max, tmp.Max);
    	ans.Min = min(ans.Min, tmp.Min);
        }
        if (r > mid) {
    	Node tmp = query(l, r, rson(x));
    	ans.sum += tmp.sum;
    	ans.Max = max(ans.Max, tmp.Max);
    	ans.Min = min(ans.Min, tmp.Min);
        }
        return ans;
    }
    
    int main() {
        while (~scanf("%d%d%d", &r, &c, &m)) {
    	build(1, r * c, 0);
    	int q, x1, y1, x2, y2, v;
    	while (m--) {
    	    scanf("%d", &q);
    	    if (q == 3) {
    		scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    		x1--; x2--;
    		int sum = 0, Max = 0, Min = INF;
    		for (int i = x1; i <= x2; i++) {
    		    Node ans = query(i * c + y1, i * c + y2, 0);
    		    sum += ans.sum;
    		    Max = max(Max, ans.Max);
    		    Min = min(Min, ans.Min);
    		}
    		printf("%d %d %d
    ", sum, Min, Max);
    	    }
    	    else {
    		scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &v);
    		x1--; x2--;
    		for (int i = x1; i <= x2; i++) {
    		    if (q == 1) add(i * c + y1, i * c + y2, v, 0);
    		    else set(i * c + y1, i * c + y2, v, 0);
    		}
    	    }
    	}
        }
        return 0;
    }


    版权声明:本文博客原创文章,博客,未经同意,不得转载。

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  • 原文地址:https://www.cnblogs.com/mfrbuaa/p/4750075.html
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