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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 |
---|---|---|---|---|---|---|---|---|---|
#707852 | #4376. Dragon slayer | hejinming983282# | AC ✓ | 146ms | 3828kb | C++23 | 3.1kb | 2024-11-03 17:50:27 | 2024-11-03 17:50:33 |
Judging History
answer
// Author : hejinming2012
#include <bits/stdc++.h>
#define int long long
#define endl '\n'
#define dbg(x) cout << #x " = " << (x) << endl
#define quickio ios::sync_with_stdio(false);
#define quickin cin.tie(0);
#define quickout cout.tie(0);
using namespace std;
inline int read() {
int now = 0, nev = 1; char c = getchar();
while(c < '0' || c > '9') { if(c == '-') nev = -1; c = getchar(); }
while(c >= '0' && c <= '9') { now = (now << 1) + (now << 3) + (c & 15); c = getchar(); }
return now * nev;
}
void write(int x) {
if(x < 0) putchar('-'), x = -x;
if(x > 9) write(x / 10);
putchar(x % 10 + '0');
}
int n, m, k, f[(1 << 15) + 5];
struct node {
int X1, Y1, X2, Y2;
} e[20];
int gl[20][20], gd[20][20];
bool vis[20][20];
int xs, ys, xt, yt;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
int dfs(int x, int y, int ex, int ey) {
vis[x][y] = 1; int res = 0;
if(x == ex && y == ey) return 1;
if(x >= 1 && gl[x][y] == 0 && vis[x - 1][y] == 0)
res |= dfs(x - 1, y, ex, ey);
if(x + 1 < n && gl[x + 1][y] == 0 && vis[x + 1][y] == 0)
res |= dfs(x + 1, y, ex, ey);
if(y >= 1 && gd[x][y] == 0 && vis[x][y - 1] == 0)
res |= dfs(x, y - 1, ex, ey);
if(y + 1 < m && gd[x][y + 1] == 0 && vis[x][y + 1] == 0)
res |= dfs(x, y + 1, ex, ey);
return res;
}
void deal(int x) {
memset(gl, 0, sizeof(gl));
memset(gd, 0, sizeof(gd));
memset(vis, 0, sizeof(vis));
for(int i = 0; i < k; i++)
if(((x >> i) & 1) == 0)
if(e[i].X1 == e[i].X2)
for(int j = e[i].Y1; j < e[i].Y2; j++)
gl[e[i].X1][j] = 1;
else if(e[i].Y1 == e[i].Y2)
for(int j = e[i].X1; j < e[i].X2; j++)
gd[j][e[i].Y1] = 1;
}
int count(int x) {
int num = 0;
while(x) num += x & 1, x >>= 1;
return num;
}
signed main() {
quickio
quickin
quickout
int T = read();
while(T--) {
n = read(), m = read(), k = read();
xs = read(), ys = read();
xt = read(), yt = read();
for(int i = 0; i < k; i++) {
e[i].X1 = read(), e[i].Y1 = read();
e[i].X2 = read(), e[i].Y2 = read();
if(e[i].X1 == e[i].X2) {
int miny = min(e[i].Y1, e[i].Y2);
int maxy = max(e[i].Y1, e[i].Y2);
e[i].Y1 = miny, e[i].Y2 = maxy;
} else if(e[i].Y1 == e[i].Y2) {
int minx = min(e[i].X1, e[i].X2);
int maxx = max(e[i].X1, e[i].X2);
e[i].X1 = minx, e[i].X2 = maxx;
}
}
memset(f, 0, sizeof(f));
int ans = k;
for(int i = 0; i < (1 << k); i++) {
if(f[i]) continue ;
deal(i), f[i] = dfs(xs, ys, xt, yt);
if(f[i]) {
ans = min(ans, count(i));
for(int j = 0; j < k; j++)
f[i | (1 << j)] |= f[i];
}
}
write(ans), putchar(10);
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 146ms
memory: 3828kb
input:
10 4 4 4 0 2 3 3 0 2 4 2 1 3 1 0 2 4 2 1 3 1 3 4 3 2 2 0 0 2 1 0 1 3 1 1 0 1 2 3 2 2 0 0 2 1 2 1 2 2 1 0 1 1 15 15 15 3 12 4 1 8 0 8 15 1 11 15 11 1 1 1 15 3 1 3 15 0 10 14 10 14 1 14 14 8 1 8 15 1 5 14 5 0 15 14 15 0 4 14 4 0 2 15 2 11 0 11 15 4 1 4 15 1 11 15 11 1 12 14 12 15 15 15 8 5 14 0 0 12 1...
output:
1 2 0 5 3 5 1 4 1 0
result:
ok 10 lines