zoj 2859 Matrix Searching

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1859

题目大意:求矩阵中任意子矩阵中的最小值.

题目思路:二维线段树.
对于这题更新相当于是单点的,所以树套树的线段树应该会比矩形树更快.
因为不懂得树套树如何对整个区间的更新(和这题没关系),所以我写二维线段树,一般都采用矩形树.

代码:

#include <stdlib.h>#include <string.h>#include <stdio.h>#include <ctype.h>#include <math.h>#include <time.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <string>#include <iostream>#include <algorithm>using namespace std;//#define ull unsigned __int64//#define ll __int64//#define ull unsigned long long//#define ll long long#define son1 New(p.xl,xm,p.yl,ym),(rt<<2)-2#define son2 New(p.xl,xm,min(ym+1,p.yr),p.yr),(rt<<2)-1#define son3 New(min(xm+1,p.xr),p.xr,p.yl,ym),rt<<2#define son4 New(min(xm+1,p.xr),p.xr,min(ym+1,p.yr),p.yr),rt<<2|1#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define middle (l+r)>>1#define MOD 1000000007#define esp (1e-8)const int INF=0x3F3F3F3F;const double DINF=10000.00;//const double pi=acos(-1.0);const int N=310;int n,m;int mmin[(N<<2)*(N<<2)];int mtx[N][N];struct node{int xl,xr,yl,yr;int xmid(){return (xl+xr)>>1;}int ymid(){return (yl+yr)>>1;}};node New(int xl,int xr,int yl,int yr){node r;r.xl=xl,r.xr=xr;r.yl=yl,r.yr=yr;return r;}void PushUp(int rt){int t1=min(mmin[(rt<<2)-2],mmin[(rt<<2)-1]);int t2=min(mmin[rt<<2],mmin[rt<<2|1]);mmin[rt]=min(t1,t2);}void Build(node p,int rt){if(p.xl==p.xr && p.yl==p.yr){mmin[rt]=mtx[p.xl][p.yl];return;}int xm=p.xmid(),ym=p.ymid();Build(son1);Build(son2);Build(son3);Build(son4);PushUp(rt);}int Query(node p,int rt,node P){if((P.xl<=p.xl&&p.xr<=P.xr)&&(P.yl<=p.yl&&p.yr<=P.yr))return mmin[rt];int xm=p.xmid(),ym=p.ymid(),ret=INF;if(P.xl<=xm){if(P.yl<=ym) ret=min(ret,Query(son1,P));if(ym<P.yr) ret=min(ret,Query(son2,P));}if(xm<P.xr){if(P.yl<=ym) ret=min(ret,Query(son3,P));if(ym<P.yr) ret=min(ret,Query(son4,P));}return ret;}int main(){//freopen("1.in","r",stdin);//freopen("1.out","w",stdout);int i,j,k,r1,c1,r2,c2;int T,cas;scanf("%d",&T);for(cas=1;cas<=T;cas++){scanf("%d",&n);for(i=1;i<=n;i++)for(j=1;j<=n;j++)scanf("%d",&mtx[i][j]);Build(New(1,n,1,n),1);scanf("%d",&m);while(m--){scanf("%d%d%d%d",&r1,&c1,&r2,&c2);if(r1>r2) swap(r1,r2);if(c1>c2) swap(c1,c2);printf("%d\n",Query(New(1,n,1,n),1,New(r1,r2,c1,c2)));}}return 0;}