QOJ.ac
QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#688504 | #5416. Fabulous Fungus Frenzy | lfyszy | WA | 1ms | 3852kb | C++14 | 4.2kb | 2024-10-30 10:15:59 | 2024-10-30 10:16:00 |
Judging History
answer
#include <bits/stdc++.h>
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define SP << " " <<
#define fish signed
using namespace std;
const int N = 20 + 10, maxc = (int)('z') + 1;
char pap[N][N];
char pre[N][N][N];
int nn[N], mm[N], n, m, k;
int have[maxc], need[N][maxc], wildcard;
//have为矩阵中每个字符的数量,need为每个预设需要的每种字符的数量
int count(int u)//计算使用当前预设所需的通配符的数量
{
int res = 0;
for(int i = '0'; i < maxc; i ++)
res += max(0, need[u][i] - have[i]);
return res;
}
bool useful(int u)//计算使用当前预设增加的通配符的数量
{
for (int i = '0'; i < maxc; i ++)
if(need[u][i] && have[i])
return true;
return false;
}
bool choose(int &now)//判断选用另一个还是使用当前预设
{
if(now > 0)
{
if(!useful(now) || wildcard < count(now))
now = 0;
}
if(now > 0) return true;
for(int i = 1; i <= k; i ++)
if(useful(i) && wildcard >= count(i))
{
now = i;
return true;
}
return false;
}
struct node {int opt, x, y;} ; vector<node> ans;
void note_swap(int x, int y, int xx, int yy)
{
if(x > xx) ans.push_back({-4, x, y});
if(x < xx) ans.push_back({-3, x, y});
if(y > yy) ans.push_back({-2, x, y});
if(y < yy) ans.push_back({-1, x, y});
swap(pap[x][y], pap[xx][yy]);
}
void note_cov(int u)
{
ans.push_back({u, 1, 1});
for(int i = 1; i <= nn[u]; i ++)
for(int j = 1; j <= mm[u]; j ++)
pap[i][j] = '*';
}
void move(char c, int gx, int gy, int sx, int sy)
{
for(int i = 1; i <= n; i ++)
for(int j = 1; j <= m; j ++)
{
if(i <= sx && j <= sy && (i < gx || i == gx && j < gy)) continue;
if(pap[i][j] == c)
{//为了不影响到已经排列好的位置,故需要视情况调整先移动哪一维
if(i < gx)
{
while(i < gx) note_swap(i, j, i + 1, j), i ++;
while(i > gx) note_swap(i, j, i - 1, j), i --;
while(j < gy) note_swap(i, j, i, j + 1), j ++;
while(j > gy) note_swap(i, j, i, j - 1), j --;
}
else
{
while(j < gy) note_swap(i, j, i, j + 1), j ++;
while(j > gy) note_swap(i, j, i, j - 1), j --;
while(i < gx) note_swap(i, j, i + 1, j), i ++;
while(i > gx) note_swap(i, j, i - 1, j), i --;
}
return ;
}
}
}
void cov(int u)
{
for(int i = 1; i <= nn[u]; i ++)
for(int j = 1; j <= mm[u]; j ++)
if(have[pre[u][i][j]])
{
move(pre[u][i][j], i, j, nn[u], mm[u]);
wildcard ++, have[pre[u][i][j]] --;
}
else move('*', i, j, nn[u], mm[u]);
note_cov(u);
}
fish main()
{
ios::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> n >> m >> k;
for(int i = 1; i <= n; i ++)
scanf("%s", pre[0][i] + 1);
nn[0] = n, mm[0] = m;//初始矩阵记为预设0
for(int i = 1; i <= n; i ++)
scanf("%s", pap[i] + 1);
for(int i = 1; i <= n; i ++)
for(int j = 1; j <= m; j ++)
have[pap[i][j]] ++;
for(int i = 1; i <= k; i ++)
{
cin >> nn[i] >> mm[i];
for(int j = 1; j <= nn[i]; j ++)
scanf("%s", pre[i][j] + 1);
}
for(int i = 0; i <= k; i ++)
for(int j = 1; j <= nn[i]; j ++)
for(int k = 1; k <= mm[i]; k ++)
need[i][pre[i][j][k]] ++;
int now = 0;
while(true)//一直替换
if(!choose(now)) break ;
else cov(now);
if(count(0) <= wildcard)//判断能否达到初始矩阵
{
if(useful(0))
{
cov(0);
ans.pop_back();
}
cout << ans.size() << "\n";
while(ans.size())
{
cout << (*ans.rbegin()).opt SP (*ans.rbegin()).x SP (*ans.rbegin()).y << "\n";
ans.pop_back();
}
}
else cout << "-1\n";
return 0;
}
詳細信息
Test #1:
score: 0
Wrong Answer
time: 1ms
memory: 3852kb
input:
3 3 1 OOO GOG BGB OOO GGG BBB 3 1 B G B
output:
0
result:
wrong answer puzzle remain unsolved