记录九度的一次TL
题目
题目描述：
在一个二维数组中，每一行都按照从左到右递增的顺序排序，每一列都按照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个整数，判断数组中是否含有该整数。
输入：
输入可能包含多个测试样例，对于每个测试案例，
输入的第一行为两个整数m和n(1<=m,n<=1000)：代表将要输入的矩阵的行数和列数。
输入的第二行包括一个整数t(1<=t<=1000000)：代表要查找的数字。
接下来的m行，每行有n个数，代表题目所给出的m行n列的矩阵(矩阵如题目描述所示，每一行都按照从左到右递增的顺序排序，每一列都按照从上到下递增的顺序排序。
输出：
对应每个测试案例，
输出”Yes”代表在二维数组中找到了数字t。
输出”No”代表在二维数组中没有找到数字t。
解答
第一反应深度优先遍历
#include <cstring>
#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <cstring>
#include <stack>
using namespace std;
enum{
WHITE = 0,
GRAY,
BLACK
};
struct map_item{
int val;
int color;
};
#define MAX_ROW 1100
#define MAX_COL 1100
struct map_item map_table[MAX_ROW][MAX_COL];
struct pos{
int row;
int col;
};
stack<struct pos> pos_store;
int m, n, key;
bool find_key;
#define CLR(a) memset(a, 0, sizeof(a))
#define CHECK_VAL(pos, kval) ( (map_table[pos.row][pos.col].val == kval ) ? (0) : (map_table[pos.row][pos.col].val > kval ? -1 : 1))
#define SET_COLOR(pos, kcolor) ( map_table[pos.row][pos.col].color = kcolor )
#define CHECK_COLOR(pos, kcolor) ( map_table[pos.row][pos.col].color == kcolor )
#define CHECK_RANGE(pos, max_row, max_col) ( !((pos.row < 1) || (pos.row > max_row) || (pos.col < 1) || (pos.col > max_col)))
#define GOTO_POS(pos) do{ \
if(CHECK_COLOR(pos, WHITE) && CHECK_RANGE(pos, m, n)){\
SET_COLOR(pos, GRAY);\
pos_store.push(pos);\
}\
}while(0)
#define LEFT_POS(pos) do{\
pos_store.pop();\
SET_COLOR(curr_pos, BLACK);\
}while(0)
#define DBG_POS(pos) do{\
cout << "pos[" << pos.row << "]["<< pos.col << "]" << endl;\
cout << "val:" << map_table[pos.row][pos.col].val << endl;\
cout << "color:" << map_table[pos.row][pos.col].color << endl;\
}while(0)
int main()
{
while(cin >> m >> n){
cin >> key;
CLR(map_table); //clean table
for(int row = 1; row <= m; row++)
for(int col = 1; col <= n; col++)
cin >> map_table[row][col].val;
struct pos curr_pos, next_pos;
curr_pos.row = 1;
curr_pos.col = 1;
find_key = false;
GOTO_POS(curr_pos);
while(!pos_store.empty())
{
if(find_key){
break; /*break! we find ans*/
}
curr_pos = pos_store.top();
LEFT_POS(curr_pos);
//DBG_POS(curr_pos);
int check_status = CHECK_VAL(curr_pos, key);
switch(check_status){
case 0:
find_key = true;
break;
case 1:
next_pos = curr_pos;
next_pos.row += 1;
GOTO_POS(next_pos);
next_pos = curr_pos;
next_pos.col += 1;
GOTO_POS(next_pos);
break;
case -1:
next_pos = curr_pos;
next_pos.row -= 1;
GOTO_POS(next_pos);
next_pos = curr_pos;
next_pos.col -= 1;
GOTO_POS(next_pos);
break;
default:
cout << "error" << endl;
exit(-1);
}
}
cout << (find_key ? "Yes":"No") << endl;
}
return 0;
}
/**************************************************************
Problem: 1384
User: xiyanxiyan10
Language: C++
Result: Time Limit Exceed
****************************************************************/
好吧，超时了
于是使用二分查找
#include <cstring>
#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <cstring>
#include <stack>
using namespace std;
#define MAX_ROW 1100
#define MAX_COL 1100
int map_table[MAX_ROW][MAX_COL];
int tot_row, tot_col, key;
int reverse_flag;
bool find_key;
#define CLR(a) memset(a, 0, sizeof(a))
/**
* @brief vist the vertex
* @param[in] i row num
* @param[in] j col num
* @note reverse_flag reverse the table
*
*/
#define GET_VAL(i, j) (reverse_flag ? map_table[j][i] : map_table[i][j])
#define REVERSE_TABLE() do{\
tot_row = tot_row^tot_col;\
tot_col = tot_row^tot_col;\
tot_row = tot_row^tot_col;\
reverse_flag ^= 1;\
}while(0)
#define INIT_TABLE() do{\
reverse_flag = 0;\
}while(0)
int main()
{
while(cin >> tot_row >> tot_col){
cin >> key;
INIT_TABLE();
for(int curr_row = 1; curr_row <= tot_row; curr_row++)
for(int curr_col = 1; curr_col <= tot_col; curr_col++)
cin >> map_table[curr_row][curr_col];
find_key = false;
/*binary_search in row, so we make sure col > row*/
if(tot_col < (tot_row )){
REVERSE_TABLE();
}
for(int curr_row = 1; (curr_row <= tot_row) && ( key > GET_VAL(curr_row, 1) ); curr_row++){
int left_col = 1;
int right_col = tot_col;
int curr_col;
if(true == find_key){
break;
}
while(left_col <= right_col){
int val;
curr_col = ((right_col - left_col) >> 1) + left_col;
//cout << " row " << curr_row << " " << curr_col << endl;
val = GET_VAL(curr_row, curr_col);
if(key == val){
find_key = true;
break;
}else if(key > val){
left_col = curr_col + 1;
}else{
right_col = curr_col - 1;
}
}
}
cout << (find_key ? "Yes":"No") << endl;
}
return 0;
}
/**************************************************************
Problem: 1384
User: xiyanxiyan10
Language: C++
Result: Time Limit Exceed
****************************************************************/
依然超时，最终代码是
#include <cstring>
#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <cstring>
#include <stack>
#include <cstdio>
using namespace std;
#define MAX_ROW 1100
#define MAX_COL 1100
bool find_key;
int tot_row, tot_col, key, tmp;
int main()
{
int input_key;
while(~scanf("%d%d%d", &tot_row, &tot_col, &key)){
find_key = false;
for(int curr_row = 1; curr_row <= tot_row; curr_row++)
for(int curr_col = 1; curr_col <= tot_col; curr_col++){
scanf("%d", &tmp);
if(tmp == key){
find_key = true;
}
}
printf(find_key ? "Yes\n":"No\n");
}
return 0;
}
总结
代码的运行时数据来自内存还是寄存器以及IO缓冲区可能对效率造成致命，需要关注。