| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 343129 | chon | 보물 찾기 (CEOI13_treasure2) | C++14 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "treasure.h"
#include <bits/stdc++.h>
//여기서 부터 Hash 구현
#define MAX_TABLE 12960000
typedef struct
{
int key[4];
int data;
}Hash;
Hash tb[MAX_TABLE];
unsigned long Hash_Key(const int* str)
{
unsigned long hash = 5381;
int c;
c = *str;
for (int i = 0; i < 4; i++)
{
hash = (((hash << 5) + hash) + c) % MAX_TABLE;
*str++;
c = *str;
}
return hash % MAX_TABLE;
}
int find(const int *key, int *data )
{
unsigned long h = Hash_Key(key);
int cnt = MAX_TABLE;
while (tb[h].key[0] != 0 && cnt--)
{
if ( tb[h].key[0]==*key && tb[h].key[1]==*key+1
&& tb[h].key[2]==*key+2 && tb[h].key[3]==*key+3 )
{
*data=tb[h].data;
return 1;
}
h = (h + 1) % MAX_TABLE;
}
return 0;
}
int add(const int *key, int *data)
{
unsigned long h = Hash_Key(key);
while (tb[h].key[0] != 0)
{
if ( tb[h].key[0]==*key && tb[h].key[1]==*key+1
&& tb[h].key[2]==*key+2 && tb[h].key[3]==*key+3)
{
return 0;
}
h = (h + 1) % MAX_TABLE;
}
tb[h].key[0]= *key;
tb[h].key[1]= *key+1;
tb[h].key[2]= *key+2;
tb[h].key[3]= *key+3;
tb[h].data= *data;
return 1;
}
using namespace std;
int ans[105][105];
int HOR[105], VER[105];
int area, included, line, ri, le;
map< tuple<int, int, int, int>, int > dp;
int askit(int r1, int c1, int r2, int c2) {
//tuple<int, int, int, int> tup(r1, c1, r2, c2);
int input[4];
input[0]=r1,input[1]=c1,input[2]=r2,input[3]=c2;
int address;
if (find(input, &address) == 1 ) return tb[address].data;
int data;
data=countTreasure(r1, c1, r2, c2);
add(input, &data);
return data;
}
int f(int v, int r, int c, int N) {
int result = 0, knew = 0, red = 0, want = 0;
if (v == 0) {
for (int i = 1; i <= r - 1; i++) {
for (int j = 1; j <= c; j++) {
knew += ans[i][j];
}
}
red = askit(1, 1, r, c);
want = red - knew;
knew = 0;
for (int j = 1; j <= c - 1; j++) knew += ans[r][j];
result = want - knew;
}
else if (v == 1) {
for (int i = 1; i <= r - 1; i++) {
for (int j = c + 1; j <= N; j++) {
knew += ans[i][j];
}
}
if (c + 1 <= N) red = askit(1, c + 1, r, N);
want = red - knew;
result = included - want;
}
else if (v == 2) {
for (int i = 1; i <= r - 1; i++) {
for (int j = c + 1; j <= N; j++) {
knew += ans[i][j];
}
}
knew = ri - knew;
if (c + 1 <= N && r + 1 <= N) red = askit(r + 1, c + 1, N, N);
want = knew - red;
result = included - want;
}
else { // D
for (int i = 1; i <= r - 1; i++) {
for (int j = 1; j <= c; j++) {
knew += ans[i][j];
}
}
knew = le - knew;
if (r + 1 <= N) red = askit(r + 1, 1, N, c);
want = knew - red;
knew = 0;
for (int j = 1; j <= c - 1; j++) knew += ans[r][j];
result = want - knew;
}
return result;
}
void findTreasure(int N) {
//int cnt = countTreasure(1, 1, N, N);
//if(cnt > 0) Report (1, 1);
area = askit(1, 1, N, N);
for (int i = 1; i <= N - 1; i++) {
if (i > (N - i)) VER[i] = askit(1, 1, i, N);
else VER[i] = area - askit(i + 1, 1, N, N);
}
for (int i = 1; i <= N - 1; i++) {
if (i > (N - i)) HOR[i] = askit(1, 1, N, i);
else HOR[i] = area - askit(1, i + 1, N, N);
}
VER[N] = HOR[N] = area;
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= N; j++) {
int below = VER[N] - VER[i];
int sup = VER[i - 1];
line = area - sup - below;
included = line;
for (int k = 1; k < j; k++) included -= ans[i][k];
ri = HOR[N] - HOR[j];
le = HOR[j];
//vector< pair<int, int> > srt;
int dx, dy;
int data[4][2];
dx = j, dy = i;
data[0][0] = dx * dy;
data[0][1] = 0;
dx = N - j, dy = i;
data[1][0] = dx * dy;
data[1][1] = 1;
dx = N - j, dy = N - i;
data[2][0] = dx * dy;
data[2][1] = 2;
dx = j, dy = N - i;
data[3][0] = dx * dy;
data[3][1] = 3;
for (int a = 0; a < 4; a++)
{
for (int b = a + 1; b < 4; b++)
{
if (data[a][0] > data[b][0])
{
int temp;
temp = data[a][0];
data[a][0] = data[b][0];
data[b][0] = temp;
temp = data[a][1];
data[a][1] = data[b][1];
data[b][1] = temp;
}
else if (data[a][0] == data[b][0] && data[a][1] > data[b][1])
{
int temp;
temp = data[a][0];
data[a][0] = data[b][0];
data[b][0] = temp;
temp = data[a][1];
data[a][1] = data[b][1];
data[b][1] = temp;
}
}
}
ans[i][j] = f(data[3][1], i, j, N);
}
}
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= N; j++) {
if (ans[i][j]) Report(i, j);
}
}