数组位置自动变为 1
Array location is becoming 1 automatically
由于我已经发布了导致超出时间限制错误的 CodeForce 问题的可能解决方案 enter link description here,因此出现了一些解决方案。尽管如此,我还是找到了另一种解决方案..
#include <stdio.h>
#include <string.h>
int greatestX = 0, greatestY = 0;
int counter = 0;
void prune_boxes(int boxes[][greatestY], int x, int y){
printf("boxes[2][0] = %d\n", boxes[2][0]);
if (y < 0)
return;
else{
//printf("x = %d y = %d\n", x, y);
if (boxes[x][y] == 0){
printf("x = %d y = %d\n", x, y);
counter++;
boxes[x][y] = 1;
}
prune_boxes(boxes, x, y - 1);
prune_boxes(boxes, x + 1, y - 1);
}
}
int main() {
int wetBoxes, i, j;
scanf("%d", &wetBoxes);
int coordinates[wetBoxes][2];
for(i = 0; i < wetBoxes; i++){
scanf("%d%d", &coordinates[i][0], &coordinates[i][1]);
if (coordinates[i][0] > greatestX)
greatestX = coordinates[i][0];
if (coordinates[i][1] > greatestY)
greatestY = coordinates[i][1];
}
int boxes[greatestY + 1][greatestY + greatestX + 1];
memset(boxes, 0, sizeof(boxes));
/*
for(i = 0; i < greatestY + 1; i++){
for (j = 0; j < greatestY + greatestX + 1; j++){
printf("%d ", boxes[i][j]);
}
printf("\n");
}
*/
for(i = 0; i < wetBoxes; i++){
//printf("value = %d\n", boxes[coordinates[i][0]][coordinates[i][1]]);
prune_boxes(boxes, coordinates[i][0], coordinates[i][1]);
}
printf("counter = %d\n", counter);
return 0;
}
现在没有导致超出时间限制,但它让我少了一个特定值的计数。
进一步调试,我发现对于 (1, 3)
的输入,代码不计算坐标 (2, 0)
。
即使进行了进一步的调试,我发现 boxes[2][0]
在我实际手动将该坐标设为 1 之前就变成了 1。
示例输出如下所示
1
1 3
boxes[2][0] = 0
x = 1 y = 3
boxes[2][0] = 1
x = 1 y = 2
boxes[2][0] = 1
x = 1 y = 1
boxes[2][0] = 1
x = 1 y = 0
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
x = 2 y = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
x = 3 y = 0
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
x = 2 y = 2
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
x = 3 y = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
x = 4 y = 0
boxes[2][0] = 1
boxes[2][0] = 1
counter = 9
如您所见,第二个递归级别的 boxes[2][0]
变为 1,但 为什么?
编辑
这部分我收到很多编译器警告
int greatestX = 0, greatestY = 0;
int counter = 0;
void prune_boxes(int boxes[][greatestY], int x, int y){
说 greatestY
不是常数。那么函数是如何知道数组大小的呢?
如果此时它接受 greatestY
作为大小,则数组将为 int boxes[][0]
,并且所有访问将为 out-of-range。这肯定会产生意想不到的结果。
并且您传递的数组无论如何都不是那些维度,但是
int boxes[greatestY + 1][greatestY + greatestX + 1];
感谢大家回答问题,看来问题出在将二维数组传递给函数。 gcc
没有抛出任何错误或警告,但 clang
给出了一些错误。因此我决定用 C++ 解决它。这是解决方案
#include <iostream>
#include <vector>
#include <algorithm>
static int counter;
std::vector<std::vector<int>> boxes;
void prune_boxes(int x, int y){
if (y < 0)
return;
else{
if (boxes[x][y] == 0){
boxes[x][y] = 1;
counter++;
}
prune_boxes(x, y - 1);
prune_boxes(x + 1, y - 1);
}
}
int main(){
int wetBoxes;
std::cin >> wetBoxes;
std::vector<int> coordinatesX;
std::vector<int> coordinatesY;
int input;
for (int i = 0; i < wetBoxes; i++){
std::cin >> input;
coordinatesX.push_back(input);
std::cin >> input;
coordinatesY.push_back(input);
}
int greatestX = *max_element(coordinatesX.begin(), coordinatesX.end());
int greatestY = *max_element(coordinatesY.begin(), coordinatesY.end());
int Y = greatestY + 1;
int X = greatestX + greatestY + 1;
boxes.resize(X);
for (int i = 0; i < X; i++){
for (int j = 0; j < Y; j++){
boxes[i].push_back(0);
}
}
for(int i = 0; i < wetBoxes; i++){
prune_boxes(coordinatesX[i], coordinatesY[i]);
std::cout << counter << std::endl;
counter = 0;
}
return 0;
}
由于我已经发布了导致超出时间限制错误的 CodeForce 问题的可能解决方案 enter link description here,因此出现了一些解决方案。尽管如此,我还是找到了另一种解决方案..
#include <stdio.h>
#include <string.h>
int greatestX = 0, greatestY = 0;
int counter = 0;
void prune_boxes(int boxes[][greatestY], int x, int y){
printf("boxes[2][0] = %d\n", boxes[2][0]);
if (y < 0)
return;
else{
//printf("x = %d y = %d\n", x, y);
if (boxes[x][y] == 0){
printf("x = %d y = %d\n", x, y);
counter++;
boxes[x][y] = 1;
}
prune_boxes(boxes, x, y - 1);
prune_boxes(boxes, x + 1, y - 1);
}
}
int main() {
int wetBoxes, i, j;
scanf("%d", &wetBoxes);
int coordinates[wetBoxes][2];
for(i = 0; i < wetBoxes; i++){
scanf("%d%d", &coordinates[i][0], &coordinates[i][1]);
if (coordinates[i][0] > greatestX)
greatestX = coordinates[i][0];
if (coordinates[i][1] > greatestY)
greatestY = coordinates[i][1];
}
int boxes[greatestY + 1][greatestY + greatestX + 1];
memset(boxes, 0, sizeof(boxes));
/*
for(i = 0; i < greatestY + 1; i++){
for (j = 0; j < greatestY + greatestX + 1; j++){
printf("%d ", boxes[i][j]);
}
printf("\n");
}
*/
for(i = 0; i < wetBoxes; i++){
//printf("value = %d\n", boxes[coordinates[i][0]][coordinates[i][1]]);
prune_boxes(boxes, coordinates[i][0], coordinates[i][1]);
}
printf("counter = %d\n", counter);
return 0;
}
现在没有导致超出时间限制,但它让我少了一个特定值的计数。
进一步调试,我发现对于 (1, 3)
的输入,代码不计算坐标 (2, 0)
。
即使进行了进一步的调试,我发现 boxes[2][0]
在我实际手动将该坐标设为 1 之前就变成了 1。
示例输出如下所示
1
1 3
boxes[2][0] = 0
x = 1 y = 3
boxes[2][0] = 1
x = 1 y = 2
boxes[2][0] = 1
x = 1 y = 1
boxes[2][0] = 1
x = 1 y = 0
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
x = 2 y = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
x = 3 y = 0
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
x = 2 y = 2
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
x = 3 y = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
boxes[2][0] = 1
x = 4 y = 0
boxes[2][0] = 1
boxes[2][0] = 1
counter = 9
如您所见,第二个递归级别的 boxes[2][0]
变为 1,但 为什么?
编辑
这部分我收到很多编译器警告
int greatestX = 0, greatestY = 0;
int counter = 0;
void prune_boxes(int boxes[][greatestY], int x, int y){
说 greatestY
不是常数。那么函数是如何知道数组大小的呢?
如果此时它接受 greatestY
作为大小,则数组将为 int boxes[][0]
,并且所有访问将为 out-of-range。这肯定会产生意想不到的结果。
并且您传递的数组无论如何都不是那些维度,但是
int boxes[greatestY + 1][greatestY + greatestX + 1];
感谢大家回答问题,看来问题出在将二维数组传递给函数。 gcc
没有抛出任何错误或警告,但 clang
给出了一些错误。因此我决定用 C++ 解决它。这是解决方案
#include <iostream>
#include <vector>
#include <algorithm>
static int counter;
std::vector<std::vector<int>> boxes;
void prune_boxes(int x, int y){
if (y < 0)
return;
else{
if (boxes[x][y] == 0){
boxes[x][y] = 1;
counter++;
}
prune_boxes(x, y - 1);
prune_boxes(x + 1, y - 1);
}
}
int main(){
int wetBoxes;
std::cin >> wetBoxes;
std::vector<int> coordinatesX;
std::vector<int> coordinatesY;
int input;
for (int i = 0; i < wetBoxes; i++){
std::cin >> input;
coordinatesX.push_back(input);
std::cin >> input;
coordinatesY.push_back(input);
}
int greatestX = *max_element(coordinatesX.begin(), coordinatesX.end());
int greatestY = *max_element(coordinatesY.begin(), coordinatesY.end());
int Y = greatestY + 1;
int X = greatestX + greatestY + 1;
boxes.resize(X);
for (int i = 0; i < X; i++){
for (int j = 0; j < Y; j++){
boxes[i].push_back(0);
}
}
for(int i = 0; i < wetBoxes; i++){
prune_boxes(coordinatesX[i], coordinatesY[i]);
std::cout << counter << std::endl;
counter = 0;
}
return 0;
}