ZOJ Monthly, February 2012 - H Matrix (ZOJ 3578)
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开始看到这题以为是二维线段树,但是二维的线段树从来没写过也不知道怎么写,昨天晚上突来灵感,可以用复杂度为c2的一个方法,每次比较两个矩阵是否相交,然后更新矩阵的值,速度很快!zoj月赛ac这题的人大部分都是这种方法吧,内存才200k。。。法克,我早该想到!!!
#include<iostream>#include<vector>#include<cstring>#include<cstdio>#include<iomanip>#include<algorithm>#include<cassert>using namespace std;const int maxn = 1011;struct zz{ int x1,y1; int x2,y2; int h;}zx[maxn];int x,y,a,b,h;int n,m,c,temp,ans;bool yes(int i,int j){ if(zx[i].y2 < zx[j].y1 || zx[i].y1 > zx[j].y2 || zx[i].x2 < zx[j].x1 || zx[i].x1 > zx[j].x2) { return false; } return true;}int main(){ while(scanf("%d%d%d",&n,&m,&c)==3) { ans = 0; for(int i=1;i<=c;i++) { // cin>>a>>b>>h>>x>>y; scanf("%d%d%d%d%d",&a,&b,&h,&x,&y); zx[i].x1 = x; zx[i].y1 = y; zx[i].x2 = x + a - 1; zx[i].y2 = y + b - 1; temp = h; for(int j=1;j<i;j++) { if(yes(i,j)) { if( h + zx[j].h > temp ) { temp = h + zx[j].h; } } } zx[i].h = temp; if(temp > ans) { ans = temp; } } printf("%d\n",ans); } return 0;}
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