hdu2888Check Corners【二维RMQ】2009 Multi-University Training Contest 9 - Host by HIT
来源:互联网 发布:广义逆矩阵求法 编辑:程序博客网 时间:2024/04/27 14:19
去年多校就出了一个裸的一维RMQ,怎么09年工大也出了啊,除了尽量少开内存,其他真的没什么,话说为毛多校的RMQ就这么简单,区域赛的就难的要死啊啊啊啊。
刚刚看一个树型dp+RMQ+二分的200+行代码,本来还想自己研究出来呢 呵呵呵
/***********hdu28882016.1.203260MS 30676K 1877B G++***********/#include <iostream>#include<cstdio>#include<cstring>using namespace std;int val[302][302];int dp[302][302][9][9];//最大值int mm[302];//二进制位数减一,使用前初始化inline int max(int x,int y){ if(x<y) return y; return x;}void initRMQ(int n,int m){for(int i = 1;i <= n;i++) for(int j = 1;j <= m;j++) dp[i][j][0][0] = val[i][j];for(int ii = 0; ii <= mm[n]; ii++) for(int jj = 0; jj <= mm[m]; jj++) if(ii+jj) for(int i = 1; i + (1<<ii) - 1 <= n;i++) for(int j = 1; j + (1<<jj) - 1 <= m;j++) { if(ii)dp[i][j][ii][jj] =max(dp[i][j][ii-1][jj],dp[i+(1<<(ii-1))][j][ii-1][jj]); else dp[i][j][ii][jj] =max(dp[i][j][ii][jj-1],dp[i][j+(1<<(jj-1))][ii][jj-1]); }}//查询矩形内的最大值(x1<=x2,y1<=y2)int rmq(int x1,int y1,int x2,int y2){ int k1 = mm[x2-x1+1]; int k2 = mm[y2-y1+1]; x2 = x2 - (1<<k1) + 1; y2 = y2 - (1<<k2) + 1; int tmp=max(max(dp[x1][y1][k1][k2],dp[x1][y2][k1][k2]),max(dp[x2][y1][k1][k2],dp[x2][y2][k1][k2])); return tmp;}int main(){ //freopen("cin.txt","r",stdin); int n,m,k; mm[0] = -1; for(int i = 1;i <= 303;i++) mm[i] = ((i&(i-1))==0)?mm[i-1]+1:mm[i-1]; while(~scanf("%d%d",&n,&m)) { for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) scanf("%d",&val[i][j]); initRMQ(n,m); scanf("%d",&k); while(k--) { int a,b,c,d; scanf("%d%d%d%d",&a,&b,&c,&d); int maxn=rmq(a,b,c,d); printf("%d ",maxn); if(maxn==val[a][b]||val[a][d]==maxn||maxn==val[c][b]||maxn==val[c][d]) puts("yes"); else puts("no"); } } return 0;}
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