ZOJ 1859 Matrix Searching(二维线段树)
2015-02-18 09:13
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http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1859
Matrix Searching
Time Limit: 10 Seconds Memory Limit: 32768 KB
Given an n*n matrix A, whose entries Ai,j are integer numbers ( 1 <= i <= n, 1 <= j <= n ). An operation FIND the minimun number in a given ssub-matrix.
Input
The first line of the input contains a single integer T , the number of test cases.
For each test case, the first line contains one integer n (1 <= n <= 300), which is the sizes of the matrix, respectively. The next n lines with n integers each gives the elements of
the matrix.
The next line contains a single integer N (1 <= N <= 1,000,000), the number of queries. The next N lines give one query on each line, with four integers r1, c1, r2, c2 (1 <= r1 <= r2
<= n, 1 <= c1 <= c2 <= n), which are the indices of the upper-left corner and lower-right corner of the sub-matrix in question.
Output
For each test case, print N lines with one number on each line, the required minimum integer in the sub-matrix.
Sample Input
1
2
2 -1
2 3
2
1 1 2 2
1 1 2 1
Sample Output
-1
2
Author: PENG, Peng
Source: ZOJ Monthly, June 2007
题意:
给出一个n*n的矩阵,有m次询问,求每次询问子矩阵中的最小值。
分析:
显然二维线段树随便乱搞搞即可了,线段树维护区域内的最小值。注意二维上的"pushup()"的写法,实际上也是要维护一棵线段树。
Matrix Searching
Time Limit: 10 Seconds Memory Limit: 32768 KB
Given an n*n matrix A, whose entries Ai,j are integer numbers ( 1 <= i <= n, 1 <= j <= n ). An operation FIND the minimun number in a given ssub-matrix.
Input
The first line of the input contains a single integer T , the number of test cases.
For each test case, the first line contains one integer n (1 <= n <= 300), which is the sizes of the matrix, respectively. The next n lines with n integers each gives the elements of
the matrix.
The next line contains a single integer N (1 <= N <= 1,000,000), the number of queries. The next N lines give one query on each line, with four integers r1, c1, r2, c2 (1 <= r1 <= r2
<= n, 1 <= c1 <= c2 <= n), which are the indices of the upper-left corner and lower-right corner of the sub-matrix in question.
Output
For each test case, print N lines with one number on each line, the required minimum integer in the sub-matrix.
Sample Input
1
2
2 -1
2 3
2
1 1 2 2
1 1 2 1
Sample Output
-1
2
Author: PENG, Peng
Source: ZOJ Monthly, June 2007
题意:
给出一个n*n的矩阵,有m次询问,求每次询问子矩阵中的最小值。
分析:
显然二维线段树随便乱搞搞即可了,线段树维护区域内的最小值。注意二维上的"pushup()"的写法,实际上也是要维护一棵线段树。
/* * * Author : fcbruce * * Date : 2014-08-24 13:03:05 * */ #include <cstdio> #include <iostream> #include <sstream> #include <cstdlib> #include <algorithm> #include <ctime> #include <cctype> #include <cmath> #include <string> #include <cstring> #include <stack> #include <queue> #include <list> #include <vector> #include <map> #include <set> #define sqr(x) ((x)*(x)) #define LL long long #define itn int #define INF 0x3f3f3f3f #define PI 3.1415926535897932384626 #define eps 1e-10 #ifdef _WIN32 #define lld "%I64d" #else #define lld "%lld" #endif #define maxm #define maxn 300 using namespace std; int n; int minv[maxn<<2][maxn<<2]; inline void pushup(int k_2d,int k) { minv[k_2d][k]=min(minv[k_2d][k*2+1],minv[k_2d][k*2+2]); } void build_1d(int k,int l,int r,int k_2d,int type) { if (r-l==1) { if (type) scanf( "%d",&minv[k_2d][k]); else minv[k_2d][k]=min(minv[k_2d*2+1][k],minv[k_2d*2+2][k]); return ; } build_1d(k*2+1,l,l+r>>1,k_2d,type); build_1d(k*2+2,l+r>>1,r,k_2d,type); pushup(k_2d,k); } void build_2d(int k,int l,int r) { if (r-l==1) build_1d(0,0,n,k,1); else { build_2d(k*2+1,l,l+r>>1); build_2d(k*2+2,l+r>>1,r); build_1d(0,0,n,k,0); } } int query_1d(int a,int b,int k,int l,int r,int k_2d) { if (b<=l || r<=a) return INF; if (a<=l && r<=b) return minv[k_2d][k]; return min(query_1d(a,b,k*2+1,l,l+r>>1,k_2d),query_1d(a,b,k*2+2,l+r>>1,r,k_2d)); } int query_2d(int a,int b,int ya,int yb,int k,int l,int r) { if (b<=l || r<=a) return INF; if (a<=l && r<=b) return query_1d(ya,yb,0,0,n,k); return min(query_2d(a,b,ya,yb,k*2+1,l,l+r>>1),query_2d(a,b,ya,yb,k*2+2,l+r>>1,r)); } int main() { #ifndef ONLINE_JUDGE freopen("/home/fcbruce/文档/code/t","r",stdin); #endif // ONLINE_JUDGE int T_T; scanf( "%d",&T_T); while (T_T--) { scanf( "%d",&n); build_2d(0,0,n); int m; scanf( "%d",&m); int x1,x2,y1,y2; while (m--) { scanf( "%d%d%d%d",&x1,&y1,&x2,&y2); x1--; y1--; printf( "%d\n",query_2d(x1,x2,y1,y2,0,0,n)); } } return 0; }
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