zoj 2589 Matrix Searching 二维线段树

给一个n*n的矩阵，给q个查询，每次给出x1, y1, x2, y2，求这个矩阵中的最小值。
代码基本上和上一题相同...
  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define pb(x) push_back(x)
  4 #define ll long long
  5 #define mk(x, y) make_pair(x, y)
  6 #define lson l, m, rt<<1
  7 #define mem(a) memset(a, 0, sizeof(a))
  8 #define rson m+1, r, rt<<1|1
  9 #define mem1(a) memset(a, -1, sizeof(a))
 10 #define mem2(a) memset(a, 0x3f, sizeof(a))
 11 #define rep(i, a, n) for(int i = a; i<n; i++)
 12 #define ull unsigned long long
 13 typedef pair<int, int> pll;
 14 const double PI = acos(-1.0);
 15 const double eps = 1e-8;
 16 const int mod = 1e9+7;
 17 const int inf = 1061109567;
 18 const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
 19 const int maxn = 305;
 20 int maxx[maxn<<2][maxn<<2], minn[maxn<<2][maxn<<2], max_ans, min_ans, n;
 21 void pushUp(int pos, int rt) {
 22     minn[pos][rt] = min(minn[pos][rt<<1], minn[pos][rt<<1|1]);
 23 }
 24 void sub_build(int sign, int pos, int l, int r, int rt) {
 25     if(l == r) {
 26         if(!sign) {
 27             scanf("%d", &minn[pos][rt]);
 28         } else {
 29             minn[pos][rt] = min(minn[pos<<1][rt], minn[pos<<1|1][rt]);
 30         }
 31         return ;
 32     }
 33     int m = l+r>>1;
 34     sub_build(sign, pos, lson);
 35     sub_build(sign, pos, rson);
 36     pushUp(pos, rt);
 37 }
 38 void build(int l, int r, int rt) {
 39     if(l == r) {
 40         sub_build(0, rt, 1, n, 1);
 41         return ;
 42     }
 43     int m = l+r>>1;
 44     build(lson);
 45     build(rson);
 46     sub_build(1, rt, 1, n, 1);
 47 }
 48 void sub_update(int sign, int pos, int y, int l, int r, int rt, int val) {
 49     if(l == r) {
 50         if(!sign) {
 51             minn[pos][rt] = val;
 52         } else {
 53             minn[pos][rt] = min(minn[pos<<1][rt], minn[pos<<1|1][rt]);
 54         }
 55         return ;
 56     }
 57     int m = l+r>>1;
 58     if(y<=m)
 59         sub_update(sign, pos, y, lson, val);
 60     else
 61         sub_update(sign, pos, y, rson, val);
 62     pushUp(pos, rt);
 63 }
 64 void update(int x, int y, int l, int r, int rt, int val) {
 65     if(l == r) {
 66         sub_update(0, rt, y, 1, n, 1, val);
 67         return ;
 68     }
 69     int m = l+r>>1;
 70     if(x<=m)
 71         update(x, y, lson, val);
 72     else
 73         update(x, y, rson, val);
 74     sub_update(1, rt, y, 1, n, 1, val);
 75 }
 76 void sub_query(int pos, int L, int R, int l, int r, int rt) {
 77     if(L<=l&&R>=r) {
 78         max_ans = max(max_ans, maxx[pos][rt]);
 79         min_ans = min(min_ans, minn[pos][rt]);
 80         return ;
 81     }
 82     int m = l+r>>1;
 83     if(L<=m)
 84         sub_query(pos, L, R, lson);
 85     if(R>m)
 86         sub_query(pos, L, R, rson);
 87 }
 88 void query(int LX, int RX, int LY, int RY, int l, int r, int rt) {
 89     if(LX<=l&&RX>=r) {
 90         sub_query(rt, LY, RY, 1, n, 1);
 91         return ;
 92     }
 93     int m = l+r>>1;
 94     if(LX<=m)
 95         query(LX, RX, LY, RY, lson);
 96     if(RX>m)
 97         query(LX, RX, LY, RY, rson);
 98 }
 99 int main()
100 {
101     int t, x, y, l, q, cnt = 1;
102     cin>>t;
103     while (t--) {
104         scanf("%d", &n);
105         build(1, n, 1);
106         cin>>q;
107         while(q--) {
108             int LX, RX, LY, RY;
109             scanf("%d%d%d%d", &LX, &LY, &RX, &RY);
110             min_ans = inf, max_ans = 0;
111             query(LX , RX, LY, RY, 1, n, 1);
112             printf("%d
", min_ans);
113         }
114     }
115 }