《算法竞赛·快冲300题 》将于2024年出版,是《算法竞赛》的辅助练习册。
所有题目放在自建的OJ New Online Judge。
用C/C++、Java、Python三种语言给出代码,以中低档题为主,适合入门、进阶。
文章目录
" 质数拼图游戏 " ,链接: http://oj.ecustacm.cn/problem.php?id=1814
题目描述
【题目描述】 给定两个nn的矩阵A和B,记C=A B(此处为矩阵乘法),存在m次询问。
每次询问C中一个子矩阵中所有数字之和。
每次询问给定a,b,c,d四个数字,表示所求子矩阵为第a行第b列到第c行第d列的子矩阵。
【输入格式】 输入第一行为n和m(1≤n≤2000,m≤50000)。
接下来n行,每行n个数字表示矩阵A。
再接下来n行,每行n个数字表示矩阵B。矩阵中每个数字不超过100。
接下来m行,每行4个数字a,b,c,d表示询问的子矩阵,(1≤a,b,c,d≤n)。
本题输入数据量大,建议使用快速读入。
【输出格式】 对于每组询问,输出一行,包含一个数字表示答案。
【输入样例】
c
3 2
1 9 8
3 2 0
1 8 3
9 8 4
0 5 15
1 9 6
1 1 3 3
2 3 1 2
【输出样例】
c
661
388
题解
如果只要求询问一个给定矩阵的子矩阵数字之和,是一个很直白的前缀和应用。
为快速得到一个矩阵的任意子矩阵的和,可以用"二维前缀和"。定义二维数组s[][], s [ i ] [ j ] s[i][j] s[i][j]表示子矩阵 [ 1 , 1 ] [ i , j ] [1, 1] ~ [i, j] [1,1] [i,j]的和。预计算出s[][]后,可以快速计算出任意的子矩阵和。如下图所示,阴影子矩阵 [ i 1 , j 1 ] [ i 2 , j 2 ] [i_1, j_1] ~ [i_2, j_2] [i1,j1] [i2,j2]的和等于:
s [ i 2 ] [ j 2 ] − s [ i 2 ] [ j 1 − 1 ] − s [ i 1 − 1 ] [ j 2 ] + s [ i 1 − 1 ] [ j 1 − 1 ] s[i_2][j_2] - s[i_2][j_1-1] - s[i_1-1][j_2] + s[i_1-1][j_1-1] s[i2][j2]−s[i2][j1−1]−s[i1−1][j2]+s[i1−1][j1−1]
其中 s [ i 1 − 1 ] [ j 1 − 1 ] s[i_1-1][ j_1-1] s[i1−1][j1−1]被减了2次,需要加回来1次。
用上述公式查询一次子矩阵和,计算量仅为O(1)。
预计算一个矩阵A的所有s[][],计算量为 n 2 n^2 n2。代码这样写:
cpp
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
cin >> A[i][j], s[i][j] = s[i-1][j]+s[i][j-1]-s[i-1][j-1]+A[i][j];//预计算s[][]
本题如果用上述方法,需要先求矩阵乘法C=A*B。但是矩阵乘法的计算量是 n 3 n^3 n3,而n≤2000,肯定超时,所以必须避免直接计算矩阵乘法。
下面分析矩阵乘法C=A*B的计算过程,看能不能利用前缀和,从而减少计算量。下图画出了矩阵乘法的细节,求C中子矩阵(a, b) ~ (c, d)的和。
(1)计算C的第b列的区间和,即 C [ a ] [ b ] + C [ a + 1 ] [ b ] + . . . + C [ c ] [ b ] C[a][b] + C[a+1][b] + ... + C[c][b] C[a][b]+C[a+1][b]+...+C[c][b]。
先看C中第b列标'*'的 C [ a ] [ b ] C[a][b] C[a][b]的计算过程,它等于A第a行乘以B第b列:
C [ a ] [ b ] = A [ a ] [ 1 ] × B [ 1 ] [ b ] + A [ a ] [ 2 ] × B [ 2 ] [ b ] + . . . + A [ a ] [ n ] × B [ n ] [ b ] C[a][b] = A[a][1]×B[1][b] + A[a][2]×B[2][b] + ... + A[a][n]×B[n][b] C[a][b]=A[a][1]×B[1][b]+A[a][2]×B[2][b]+...+A[a][n]×B[n][b]
同理,C中第b列的其他坐标的计算过程是:
C [ a + 1 ] [ b ] = A [ a + 1 ] [ 1 ] × B [ 1 ] [ b ] + A [ a + 1 ] [ 2 ] × B [ 2 ] [ b ] + . . . + A [ a + 1 ] [ n ] × B [ n ] [ b ] C[a+1][b] = A[a+1][1]×B[1][b] + A[a+1][2]×B[2][b] + ...+ A[a+1][n]×B[n][b] C[a+1][b]=A[a+1][1]×B[1][b]+A[a+1][2]×B[2][b]+...+A[a+1][n]×B[n][b]
...
C [ c ] [ b ] = A [ c ] [ 1 ] × B [ 1 ] [ b ] + A [ c ] [ 2 ] × B [ 2 ] [ b ] + . . . + A [ c ] [ n ] × B [ n ] [ b ] C[c][b] = A[c][1]×B[1][b] + A[c][2]×B[2][b] + ... +A[c][n]×B[n][b] C[c][b]=A[c][1]×B[1][b]+A[c][2]×B[2][b]+...+A[c][n]×B[n][b] (式3-1)
把(式3-1)上下相加得C的子矩阵第b列的区间和:
C [ a ] [ b ] + C [ a + 1 ] [ b ] + . . . + C [ c ] [ b ] C[a][b] + C[a+1][b] + ... + C[c][b] C[a][b]+C[a+1][b]+...+C[c][b]
= ( A [ a ] [ 1 ] + A [ a + 1 ] [ 1 ] + . . . + A [ c ] [ 1 ] ) × B [ 1 ] [ b ] + = (A[a][1]+A[a+1][1] + ... + A[c][1])×B[1][b] + =(A[a][1]+A[a+1][1]+...+A[c][1])×B[1][b]+
( A [ a ] [ 2 ] + A [ a + 1 ] [ 2 ] + . . . + A [ c ] [ 2 ] ) × B [ 2 ] [ b ] + (A[a][2]+A[a+1][2] + ... + A[c][2])×B[2][b] + (A[a][2]+A[a+1][2]+...+A[c][2])×B[2][b]+
...
( A [ a ] [ n ] + A [ a + 1 ] [ n ] + . . . + A [ c ] [ n ] ) × B [ n ] [ b ] (A[a][n]+A[a+1][n] + ... + A[c][n])×B[n][b] (A[a][n]+A[a+1][n]+...+A[c][n])×B[n][b] (式3-2)
式中的 A[a][1]+A[a+1][1]+...+A[c][1]正好是A的第1列的区间和,A[a][2]+A[a+1][2]+...+A[c][2]是第2列的区间和,...,等等。
记s1[][j]为A的第j列的前缀和,有:
A [ a ] [ 1 ] + A [ a + 1 ] [ 1 ] + . . . + A [ c ] [ 1 ] = s 1 [ c ] [ 1 ] − s 1 [ a − 1 ] [ 1 ] A[a][1]+A[a+1][1]+...+A[c][1] = s1[c][1] - s1[a-1][1] A[a][1]+A[a+1][1]+...+A[c][1]=s1[c][1]−s1[a−1][1]
A [ a ] [ 2 ] + A [ a + 1 ] [ 2 ] + . . . + A [ c ] [ 2 ] = s 1 [ c ] [ 2 ] − s 1 [ a − 1 ] [ 2 ] A[a][2]+A[a+1][2]+...+A[c][2] = s1[c][2] - s1[a-1][2] A[a][2]+A[a+1][2]+...+A[c][2]=s1[c][2]−s1[a−1][2]
...
A [ a ] [ n ] + A [ a + 1 ] [ n ] + . . . + A [ c ] [ n ] = s 1 [ c ] [ n ] − s 1 [ a − 1 ] [ n ] A[a][n]+A[a+1][n] + ... + A[c][n]=s1[c][n] - s1[a-1][n] A[a][n]+A[a+1][n]+...+A[c][n]=s1[c][n]−s1[a−1][n]
则C的子矩阵第b列的区间和(式3-2)简化为:
C [ a c ] [ b ] C[a~c][b] C[a c][b]
= ( s 1 [ c ] [ 1 ] − s 1 [ a − 1 ] [ 1 ] ) × B [ 1 ] [ b ] + ( s 1 [ c ] [ 2 ] − s 1 [ a − 1 ] [ 2 ] ) × B [ 2 ] [ b ] + . . . + s 1 [ c ] [ n ] − s 1 [ a − 1 ] [ n ] × B [ n ] [ b ] = (s1[c][1] - s1[a-1][1])×B[1][b] + (s1[c][2] - s1[a-1][2])×B[2][b] + ...+s1[c][n] - s1[a-1][n]×B[n][b] =(s1[c][1]−s1[a−1][1])×B[1][b]+(s1[c][2]−s1[a−1][2])×B[2][b]+...+s1[c][n]−s1[a−1][n]×B[n][b]
(2)计算C的子矩阵的和,即把C的第b列、b+1列、...、d列相加。根据(1)的讨论,有:
C [ a c ] [ b ] + C [ a c ] [ b + 1 ] + . . . + C [ a c ] [ d ] C[a~c][b] + C[a~c][b+1] + ... + C[a~c][d] C[a c][b]+C[a c][b+1]+...+C[a c][d]
= ( s 1 [ c ] [ 1 ] − s 1 [ a − 1 ] [ 1 ] ) × B [ 1 ] [ b ] + ( s 1 [ c ] [ 2 ] − s 1 [ a − 1 ] [ 2 ] ) × B [ 2 ] [ b ] + . . . = (s1[c][1] - s1[a-1][1])×B[1][b] + (s1[c][2] - s1[a-1][2])×B[2][b] + ... =(s1[c][1]−s1[a−1][1])×B[1][b]+(s1[c][2]−s1[a−1][2])×B[2][b]+...
( s 1 [ c ] [ 1 ] − s 1 [ a − 1 ] [ 1 ] ) × B [ 1 ] [ b + 1 ] + ( s 1 [ c ] [ 2 ] − s 1 [ a − 1 ] [ 2 ] ) × B [ 2 ] [ b + 1 ] + . . . (s1[c][1] - s1[a-1][1])×B[1][b+1] + (s1[c][2] - s1[a-1][2])×B[2][b+1] + ... (s1[c][1]−s1[a−1][1])×B[1][b+1]+(s1[c][2]−s1[a−1][2])×B[2][b+1]+...
( s 1 [ c ] [ 1 ] − s 1 [ a − 1 ] [ 1 ] ) × B [ 1 ] [ b + 2 ] + ( s 1 [ c ] [ 2 ] − s 1 [ a − 1 ] [ 2 ] ) × B [ 2 ] [ b + 2 ] + . . . (s1[c][1] - s1[a-1][1])×B[1][b+2] + (s1[c][2] - s1[a-1][2])×B[2][b+2] + ... (s1[c][1]−s1[a−1][1])×B[1][b+2]+(s1[c][2]−s1[a−1][2])×B[2][b+2]+...
...
( s 1 [ c ] [ 1 ] − s 1 [ a − 1 ] [ 1 ] ) × B [ 1 ] [ d ] + ( s 1 [ c ] [ 2 ] − s 1 [ a − 1 ] [ 2 ] ) × B [ 2 ] [ d ] + . . . (s1[c][1] - s1[a-1][1])×B[1][d] + (s1[c][2] - s1[a-1][2])×B[2][d] + ... (s1[c][1]−s1[a−1][1])×B[1][d]+(s1[c][2]−s1[a−1][2])×B[2][d]+... (式3-3)
把(式3-3)上下相加,得:
C [ a c ] [ b ] + C [ a c ] [ b + 1 ] + . . . + C [ a c ] [ d ] C[a~c][b] + C[a~c][b+1] + ... + C[a~c][d] C[a c][b]+C[a c][b+1]+...+C[a c][d]
= ( s 1 [ c ] [ 1 ] − s 1 [ a − 1 ] [ 1 ] ) × ( B [ 1 ] [ b ] + B [ 1 ] [ b + 1 ] + . . . + B [ 1 ] [ d ] ) + = (s1[c][1] - s1[a-1][1]) × (B[1][b]+B[1][b+1] + ...+ B[1][d]) + =(s1[c][1]−s1[a−1][1])×(B[1][b]+B[1][b+1]+...+B[1][d])+
( s 1 [ c ] [ 2 ] − s 1 [ a − 1 ] [ 2 ] ) × ( B [ 2 ] [ b ] + B [ 2 ] [ b + 1 ] + . . . + B [ 2 ] [ d ] ) + (s1[c][2] - s1[a-1][2]) × (B[2][b]+B[2][b+1] + ... + B[2][d]) + (s1[c][2]−s1[a−1][2])×(B[2][b]+B[2][b+1]+...+B[2][d])+
...
( s 1 [ c ] [ n ] − s 1 [ a − 1 ] [ n ] ) × ( B [ n ] [ b ] + B [ n ] [ b + 1 ] + . . . + B [ n ] [ d ] ) (s1[c][n] - s1[a-1][n]) × (B[n][b]+B[n][b+1] + ... + B[n][d]) (s1[c][n]−s1[a−1][n])×(B[n][b]+B[n][b+1]+...+B[n][d]) (式3-4)
记s2[i][]为B的第i行的前缀和,有:
B [ 1 ] [ b ] + B [ 1 ] [ b + 1 ] + . . . + B [ 1 ] [ d ] = s 2 [ 1 ] [ d ] − s 2 [ 1 ] [ b − 1 ] B[1][b]+B[1][b+1] + ...+ B[1][d] = s2[1][d] - s2[1][b-1] B[1][b]+B[1][b+1]+...+B[1][d]=s2[1][d]−s2[1][b−1]
B [ 2 ] [ b ] + B [ 2 ] [ b + 1 ] + . . . + B [ 2 ] [ d ] = s 2 [ 2 ] [ d ] − s 2 [ 2 ] [ b − 1 ] B[2][b]+B[2][b+1] + ...+ B[2][d] = s2[2][d] - s2[2][b-1] B[2][b]+B[2][b+1]+...+B[2][d]=s2[2][d]−s2[2][b−1]
...
B [ n ] [ b ] + B [ n ] [ b + 1 ] + . . . + B [ n ] [ d ] = s 2 [ n ] [ d ] − s 2 [ n ] [ b − 1 ] B[n][b]+B[n][b+1] + ... + B[n][d]=s2[n][d] - s2[n][b-1] B[n][b]+B[n][b+1]+...+B[n][d]=s2[n][d]−s2[n][b−1]
则(式3-4)改写为:
C [ a c ] [ b ] + C [ a c ] [ b + 1 ] + . . . + C [ a c ] [ d ] C[a~c][b] + C[a~c][b+1] + ... + C[a~c][d] C[a c][b]+C[a c][b+1]+...+C[a c][d]
= ( s 1 [ c ] [ 1 ] − s 1 [ a − 1 ] [ 1 ] ) × ( s 2 [ 1 ] [ d ] − s 2 [ 1 ] [ b − 1 ] ) + = (s1[c][1] - s1[a-1][1]) × ( s2[1][d] - s2[1][b-1]) + =(s1[c][1]−s1[a−1][1])×(s2[1][d]−s2[1][b−1])+
( s 1 [ c ] [ 2 ] − s 1 [ a − 1 ] [ 2 ] ) × ( s 2 [ 2 ] [ d ] − s 2 [ 2 ] [ b − 1 ] ) + (s1[c][2] - s1[a-1][2]) × ( s2[2][d] - s2[2][b-1]) + (s1[c][2]−s1[a−1][2])×(s2[2][d]−s2[2][b−1])+
...
( s 1 [ c ] [ n ] − s 1 [ a − 1 ] [ n ] ) × ( s 2 [ n ] [ d ] − s 2 [ n ] [ b − 1 ] ) (s1[c][n] - s1[a-1][n]) × ( s2[n][d] - s2[n][b-1]) (s1[c][n]−s1[a−1][n])×(s2[n][d]−s2[n][b−1])
这是最后的式子,每一行是两个区间和的乘法,共n行,有n次乘法计算。
总计算量是多少?(1)预计算s1[][]和s2[][],是 O ( n 2 ) O(n^2) O(n2)的;(2)查询m次子矩阵和,每次有n次乘法计算,是 O ( m n ) O(mn) O(mn)的;(3)总计算量等于 O ( n 2 ) + O ( m n ) O(n^2) + O(mn) O(n2)+O(mn),刚好通过测试。
【重点】 前缀和,矩阵计算 。
C++代码
题目中提到"本题输入数据量大,建议使用快速读入"。
C++的标准输入输出函数是cin/cout、scanf/printf,在默认情况下,cin/cout比scanf/printf慢得多。在需要大量输入输出的场合,一般用scanf、printf就可以。如果还要提高速度,输入用getchar(),输出用putchar(),它们更快。
自己写一个快读函数read(),用到getchar()。getchar()的功能是读1 byte的数据,按char类型读入。下面代码中的read()是整数输入的快读模板,用getchar()读入每个字符,然后转成数字。例如输入"245",用getchar()分3次读入'2'、'4'、'5',然后组合成数字"345"。注意可能有负数,所以需要判断'-'号。
自己写一个快写函数write(),用到putchar()。putchar()的功能是输出一个字符,当需要输出一个数时,把它的每一位转成字符,然后用putchar()输出。
cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2010;
int n,m,a,b,c,d;
int A[N][N],B[N][N],s1[N][N],s2[N][N];
inline int read(int &x) { //快读int型整数。如果需要读long long,把int改成long long
x = 0;
int w = 1;//w:判断正负号
char ch = 0;
while (ch < '0' || ch > '9') { //读字符
if (ch == '-') w = -1; //这是一个负整数数
ch = getchar(); //读一个字符
}
while (ch >= '0' && ch <= '9') { //读数字
x = x * 10 + (ch - '0');
ch = getchar();
}
return x = x * w;
}
void write(ll x) { //快写long long型整数
if (x < 0) { // 判断正负。如果是负数,输出负号
putchar('-');
x = -x; //记得把负数变正,方便下面输出数字
}
if (x > 9) write(x / 10); // 递归,将除最后一位外的其他部分放到递归中输出
putchar(x % 10 + '0'); // 已经输出(递归)完 x 末位前的所有数字,输出末位
}
ll query(int a,int b,int c,int d){ //C=A*B,计算C的子矩阵和
ll ans = 0;
for(int k=1;k<=n;k++){
ll ans1 = s1[c][k] - s1[a-1][k];
ll ans2 = s2[k][d] - s2[k][b-1];
ans += ans1*ans2;
}
return ans;
}
int main(){
read(n),read(m);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
read(A[i][j]), s1[i][j] = s1[i-1][j]+A[i][j]; //输入A。s1[][j]是第j列的前缀和
//read(A[i][j])等于scanf("%d",&A[i][j])或cin>>A[i][j]
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
read(B[i][j]), s2[i][j] = s2[i][j-1]+B[i][j]; //输入B。s2[i][]是第i行的前缀和
while(m--){
read(a),read(b),read(c),read(d); //等于scanf("%d%d%d%d",&a,&b,&c,&d);
if(a > c) swap(a, c); //可能存在a>c、b>d的情况
if(b > d) swap(b, d);
ll ans = query(a,b,c,d);
write(ans); putchar('\n'); //等于printf("%lld\n",query(a,b,c,d));
}
return 0;
}
Java代码
java
import java.util.*;
import java.io.*;
class Main {
static FastReader scanner = new FastReader();
static int N = 2010;
static int n, m, a, b, c, d;
static int[][] A = new int[N][N], B = new int[N][N], s1 = new int[N][N], s2 = new int[N][N];
public static void main(String[] args) throws IOException {
n = scanner.nextInt();
m = scanner.nextInt();
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) {
A[i][j] = scanner.nextInt();;
s1[i][j] = s1[i - 1][j] + A[i][j];
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) {
B[i][j] = scanner.nextInt();;
s2[i][j] = s2[i][j - 1] + B[i][j];
}
while (m-- > 0) {
a = scanner.nextInt();
b = scanner.nextInt();
c = scanner.nextInt();
d = scanner.nextInt();
if (a > c) { int temp = a; a = c; c = temp; }
if (b > d) { int temp = b; b = d; d = temp; }
long ans = query(a, b, c, d);
System.out.println(ans);
}
}
static long query(int a, int b, int c, int d) {
long ans = 0;
for (int k = 1; k <= n; k++) {
long ans1 = s1[c][k] - s1[a - 1][k];
long ans2 = s2[k][d] - s2[k][b - 1];
ans += ans1 * ans2;
}
return ans;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {st = new StringTokenizer(br.readLine());}
catch (IOException e) { e.printStackTrace(); }
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble() { return Double.parseDouble(next()); }
String nextLine() {
String str = "";
try { str = br.readLine(); }
catch (IOException e) { e.printStackTrace(); }
return str;
}
}
}
Python代码
python
#pypy
import sys
input = sys.stdin.readline
def query(a, b, c, d):
ans = 0
for k in range(1, n+1):
ans1 = s1[c][k] - s1[a-1][k]
ans2 = s2[k][d] - s2[k][b-1]
ans += ans1 * ans2
return ans
n, m = list(map(int, input().split()))
N = n+1
s1 = [[0] * N for _ in range(N)]
s2 = [[0] * N for _ in range(N)]
A = [0] * N
B = [0] * N
for i in range(1, n+1):
A[i] = [0] + list(map(int, input().split()))
for j in range(1, n+1): s1[i][j] = s1[i-1][j] + A[i][j]
for i in range(1, n+1):
B[i] = [0] + list(map(int, input().split()))
for j in range(1, n+1): s2[i][j] = s2[i][j-1] + B[i][j]
for _ in range(m):
a, b, c, d = list(map(int, input().split()))
if a > c: a, c = c, a
if b > d: b, d = d, b
sys.stdout.write(str(query(a, b, c, d)) + '\n')