Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 264 Accepted Submission(s): 106
题目链接:HDU 4845
很显然用三元组$(x,y,state)$表示当前x、y坐标,获取的钥匙情况,而且这样是可以唯一确定状态的,然后钥匙获取情况用状态压缩表示即可
代码:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 16;
struct info
{
int x, y;
int keyst, step;
info() {}
info(int _x, int _y, int _keyst, int _step): x(_x), y(_y), keyst(_keyst), step(_step) {}
};
int dir[4][2] = {{1, 0}, {0, 1}, { -1, 0}, {0, -1}};
int door[N][N][N][N];
int key[N][N];
int d[N][N][1 << 11];
bool vis[N][N][1 << 11];
int n, m, p, k, s;
void init()
{
CLR(door, 0);
CLR(key, 0);
CLR(d, INF);
CLR(vis, false);
}
inline bool check(const int &x, const int &y)
{
return x >= 1 && x <= n && y >= 1 && y <= m;
}
void bfs()
{
queue<info>Q;
info S(1, 1, 0 | key[1][1], 0);
Q.push(S);
vis[S.x][S.y][S.keyst] = true;
d[S.x][S.y][S.keyst] = 1;
while (!Q.empty())
{
info u = Q.front();
Q.pop();
for (int i = 0; i < 4; ++i)
{
int vx = u.x + dir[i][0];
int vy = u.y + dir[i][1];
if (!check(vx, vy))
continue;
int need = door[u.x][u.y][vx][vy];
if (need == -1)
continue;
if ((need == 0 || (u.keyst & (1 << need))))
{
int vkeyst = u.keyst | (key[vx][vy]);
if (!vis[vx][vy][vkeyst])
{
vis[vx][vy][vkeyst] = true;
d[vx][vy][vkeyst] = u.step + 1;
Q.push(info(vx, vy, vkeyst, d[vx][vy][vkeyst]));
}
}
}
}
}
int main(void)
{
int i;
while (~scanf("%d%d%d", &n, &m, &p))
{
init();
scanf("%d", &k);
for (i = 0; i < k; ++i)
{
int x1, x2, y1, y2, g;
scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &g);
if (!g)
door[x1][y1][x2][y2] = door[x2][y2][x1][y1] = -1;
else
door[x1][y1][x2][y2] = door[x2][y2][x1][y1] = g;
}
scanf("%d", &s);
for (i = 0; i < s; ++i)
{
int x1, y1, qi;
scanf("%d%d%d", &x1, &y1, &qi);
key[x1][y1] |= (1 << qi);
}
bfs();
int ans = INF;
int Alst = (1 << (p + 1));
for (i = 0; i < Alst; ++i)
if (d[n][m][i] < ans)
ans = d[n][m][i];
printf("%d\n", ans == INF ? -1 : ans);
}
return 0;
}
原文:http://www.cnblogs.com/Blackops/p/7217062.html