RMQ (Range Minimum/Maximum Query) and LCA(Lowest Common Ancestor)
学后缀数组的时候遇到这个问题,不会,特地去找了资料…
发现农夫三拳的一篇强大翻译阐述了LCA和RMQ的关系,还有郭华阳 《RMQ与LCA问题》
区间最值问题,强大就强大在O(1)的询问
寻思着kth number能否由这个变形得到?
其他也没啥好探讨的…
献上模板和题目
3264 Balanced Lineup 一维RMQ

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#define M 50001
int val[M];
int Max[20][M];
int Min[20][M];
int idx[M];
void initRMQ(int n) {
	idx[0] = -1;
	for(int i = 1; i <= n ; i ++) {
		idx[i] = (i&(i-1)) ? idx[i-1] : idx[i-1] + 1;
		Min[0][i] = Max[0][i] = val[i];
	}
	for(int i = 1; i <= idx[n] ; i ++) {
		int limit = n + 1 - (1<<i);
		for(int j = 1; j <= limit ; j ++) {
			Min[i][j] = min(Min[i-1][j] , Min[i-1][j+(1<<i>>1)]);
			Max[i][j] = max(Max[i-1][j] , Max[i-1][j+(1<<i>>1)]);
		}
	}
}
int getval(int a,int b) {
	int t = idx[b-a+1];
	b -= (1<<t) - 1;
	return max(Max[t][a] , Max[t][b]) - min(Min[t][a] , Min[t][b]);//返回最大值减最小值
}
int main() {
	int n , m;
	scanf("%d%d",&n,&m);
	for(int i = 1; i <= n ; i ++) {//下标要从1开始
		SS(val[i]);
	}
	initRMQ(n);
	while(m --) {
		int a , b ;
		scanf("%d%d",&a,&b);
		printf("%d\n",getval(a,b));
	}
	return 0;
}

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hdu2888 Check Corners 二维RMQ

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//复杂度n*m*log(n)*log(m)
#define M 301
int val[M][M];
int Max[9][9][M][M];
int idx[M];
 
void initRMQ(int n , int m) {
	for(int i = 1; i <= n ; i ++) {
		for(int j = 1 ; j <= m ; j ++) {
			Max[0][0][i][j] = val[i][j];
		}
	}
	for(int i = 0; i <= idx[n] ; i ++) {
		int limit1 = n + 1 - (1<<i);
		for(int j = 0 ; j <= idx[m] ; j ++) {
			if(!i && !j)	continue;
			int limit2 = m + 1 - (1<<j);
			for(int ii = 1; ii <= limit1 ; ii ++) {
				for(int jj = 1; jj <= limit2 ; jj ++) {
					if(i)	Max[i][j][ii][jj] = max( Max[i-1][j][ii+(1<<i>>1)][jj] , Max[i-1][j][ii][jj] );
					else	Max[i][j][ii][jj] = max( Max[i][j-1][ii][jj] , Max[i][j-1][ii][jj+(1<<j>>1)] );
				}
			}
		}
	}
}
int query(int a,int b,int c,int d) {
	int n = idx[c-a+1]	,	m = idx[d-b+1];
	c -= (1<<n) - 1;		d -= (1<<m) - 1;
	return max(max(Max[n][m][a][b],Max[n][m][a][d]),max(Max[n][m][c][b],Max[n][m][c][d]));
}
int main() {
	idx[0] = -1;
	for(int i = 1; i <= 300 ; i ++) {
		idx[i] = (i&(i-1)) ? idx[i-1] : idx[i-1] + 1;//放在外边计算
	}
	int n , m , Q;
	while(~scanf("%d%d",&n,&m)) {
		for(int i = 1; i <= n ; i ++) {
			for(int j = 1; j <= m ; j ++) {
				SS(val[i][j]);
			}
		}
		initRMQ(n,m);
		SS(Q);
		while(Q --) {
			int a,b,c,d;
			scanf("%d%d%d%d",&a,&b,&c,&d);
			if(a > c) swap(a,c);
			if(b > d) swap(b,d);
			int key = query(a,b,c,d);
			printf("%d ",key);
			if(key == val[a][b] || key == val[a][d] || key == val[c][b] || key == val[c][d]) {
				puts("yes");
			} else {
				puts("no");
			}
		}
	}
}

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