网络流24题 之十四 孤岛营救问题 分层图

题目大意：一张网格图，上面有一些点可能有某种钥匙。节点和节点之间可能有门。有些门须要特定的钥匙就能够通过，有些不管怎样都过不去。求从(1,1)開始到(m,n)的最短时间。

思路：分层图+状态压缩。

f[i][j][k]，当中i和j描写叙述的是当前所在的位置。k是压缩了的当前有哪些钥匙（因为钥匙的数量<=10，所以全部的状态1<<10的空间内就能够搞定）。然后向四个方向更新的时候推断能否经过门。

CODE：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 20
#define INF 0x3f3f3f3f
using namespace std;
const int dx[] = {0,1,-1,0,0};
const int dy[] = {0,0,0,1,-1};

struct Complex{
int x,y,status;

Complex(int _,int __,int ___):x(_),y(__),status(___) {}
Complex() {}
};

int m,n,p,doors,keys;
int f[MAX][MAX][1 << 11];
bool v[MAX][MAX][1 << 11];
int map[MAX][MAX][MAX][MAX];
int key[MAX][MAX];

void SPFA();
inline bool Accelerator(int x1,int y1,int x2,int y2,int status);

int main()
{
cin >> m >> n >> p >> doors;
memset(map,-1,sizeof(map));
for(int a,b,c,d,x,i = 1;i <= doors; ++i) {
scanf("%d%d%d%d%d",&a,&b,&c,&d,&x);
map[a][b][c][d] = map[c][d][a][b] = x;
}
cin >> keys;
for(int x,y,z,i = 1;i <= keys; ++i) {
scanf("%d%d%d",&x,&y,&z);
key[x][y] |= 1 << z;
}
SPFA();
int ans = INF;
for(int i = 0;i < (1 << 11); ++i)
ans = min(ans,f[m]
[i]);
if(ans == INF)	ans = -1;
cout << ans << endl;
return 0;
}

void SPFA()
{
memset(f,0x3f,sizeof(f));
f[1][1][0] = 0;
static queue<Complex> q;
while(!q.empty())	q.pop();
q.push(Complex(1,1,0));
while(!q.empty()) {
Complex now = q.front(); q.pop();
v[now.x][now.y][now.status] = false;
int _status = now.status;
if(key[now.x][now.y])	_status |= key[now.x][now.y];
for(int i = 1;i <= 4; ++i) {
int fx = now.x + dx[i];
int fy = now.y + dy[i];
if(!fx || !fy || fx > m || fy > n)	continue;
if(Accelerator(now.x,now.y,fx,fy,_status))
if(f[fx][fy][_status] > f[now.x][now.y][now.status] + 1) {
f[fx][fy][_status] = f[now.x][now.y][now.status] + 1;
if(!v[fx][fy][_status])
v[fx][fy][_status] = true,q.push(Complex(fx,fy,_status));
}
}
}
}

inline bool Accelerator(int x1,int y1,int x2,int y2,int status)
{
int need_key = map[x1][y1][x2][y2];
if(!need_key)	return false;
if(need_key == -1)	return true;
return (status >> need_key)&1;
}