UVA11992 - Fast Matrix Operations(段樹部分的變化)

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UVA11992 - Fast Matrix Operations(線段樹區間改動)

題目連結

題目大意：給你個r*c的矩陣，初始化為0。
然後給你三種操作：
1 x1, y1, x2, y2, v 把由x1，y1, x2, y2構成的子矩陣裡的每一個元素都加上v。
2 x1, y1, x2, y2, v 把這個子矩陣的每一個元素都改動為v。
3 x1, y1, x2, y2 查詢這個子矩陣的元素之和，和這些元素的最大值和最小值。

解題思路：由於矩陣的行最多20行，所以能夠將這個矩陣的元素構建一棵線段樹，每一個節點都有三個附加資訊：sum，Max_num, Min_num.然後改動查詢的時候就每行每行的來做。注意：每次set的時候都要將這個節點的add清理掉。由於add已經是不須要的了。

然後每次的pushdown都要先處理set再處理add。pushdown處理的是這個節點的子節點。然後通過pushup就能夠更新這個節點的附加資訊。

代碼:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define lson(x) ((x)<<1)#define rson(x) (((x)<<1) + 1)const int N = 1e6 + 5;const int INF = 0x3f3f3f3f;struct Item {    int sum, Max_num, Min_num;    Item (int sum = 0, int Max_num = -INF, int Min_num = INF) {        this->set(sum, Max_num, Min_num);    }    void set (int sum, int Max_num, int Min_num) {        this->sum = sum;        this->Max_num = Max_num;        this->Min_num  = Min_num;    }};struct Node {    Item v;    int l, r;    int addv, setv;    void set (int l, int r, int addv = 0, int setv = -1) {        this->l = l;        this->r = r;        this->addv = addv;        this->setv = setv;        }}node[4 * N];Item get_item (Item a, Item b) {    return Item(a.sum + b.sum, max(a.Max_num, b.Max_num), min(a.Min_num, b.Min_num));}void Node_add(int u, int addv) {    node[u].addv += addv;    node[u].v.sum += addv * (node[u].r - node[u].l + 1);    node[u].v.Max_num += addv;    node[u].v.Min_num += addv;}void Node_set(int u, int setv) {    node[u].setv = setv;    node[u].addv = 0;    node[u].v.sum = setv * (node[u].r - node[u].l + 1);    node[u].v.Max_num = node[u].v.Min_num = setv;}void pushup (int u) {    node[u].set(node[lson(u)].l, node[rson(u)].r);    node[u].v = get_item(node[lson(u)].v, node[rson(u)].v);}void pushdown(int u) {    if (node[u].setv >= 0) {        Node_set(lson(u), node[u].setv);        Node_set(rson(u), node[u].setv);        node[u].setv = -1;    }    if (node[u].addv) {        Node_add(lson(u), node[u].addv);        Node_add(rson(u), node[u].addv);        node[u].addv = 0;        }}void Build (int u, int l, int r) {    if (l == r) {         node[u].set(l, r);        node[u].v.set(0, 0, 0);    } else {        int m = (l + r) / 2;        Build(lson(u), l, m);        Build(rson(u), m + 1, r);        pushup(u);    }}void Add (int u, int l, int r, int addv) {    if (node[u].l >= l && node[u].r <= r)        Node_add(u, addv);    else {        pushdown(u);        int m = (node[u].l + node[u].r) / 2;        if (l <= m)            Add (lson(u), l, r, addv);        if (r > m)            Add (rson(u), l, r, addv);        pushup(u);    }}void Set (int u, int l, int r, int setv) {    if (node[u].l >= l && node[u].r <= r)        Node_set(u, setv);    else {        pushdown(u);        int m = (node[u].l + node[u].r) / 2;        if (l <= m)            Set (lson(u), l, r, setv);        if (r > m)            Set (rson(u), l, r, setv);        pushup(u);    }}Item Query (int u, int l, int r) {    if (node[u].l >= l && node[u].r <= r)         return node[u].v;    else {        Item ans;        pushdown(u);        int m = (node[u].l + node[u].r) / 2;        if (l <= m)            ans = get_item (ans, Query(lson(u), l, r));        if (r > m)            ans = get_item (ans, Query(rson(u), l, r));            pushup(u);        return ans;    }}int main () {    int r, c, m;    int type;    int x1, y1, x2, y2, v;    while (scanf ("%d%d%d", &r, &c, &m) != EOF) {        Build(1, 1, r * c);        while (m--) {            scanf ("%d", &type);            if (type == 1) {                scanf ("%d%d%d%d%d", &x1, &y1, &x2, &y2, &v);                for (int i = x1 - 1; i < x2; i++)                    Add(1, i * c + y1, i * c + y2, v);            } else if (type == 2) {                scanf ("%d%d%d%d%d", &x1, &y1, &x2, &y2, &v);                for (int i = x1 - 1; i < x2; i++)                    Set(1, i * c + y1, i * c + y2, v);            } else {                scanf ("%d%d%d%d", &x1, &y1, &x2, &y2);                Item ans;                for (int i = x1 - 1; i < x2; i++)                    ans = get_item(ans, Query(1, i * c + y1, i * c + y2));                printf ("%d %d %d\n", ans.sum, ans.Min_num, ans.Max_num);            }        }    }    return 0;}

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UVA11992 - Fast Matrix Operations(段樹部分的變化)