分析
这题直接状压BFS也可以，不过可以把BFS改成SPFA，进一步优化
显然要维护三个状态:s,x,y，分别表示钥匙状态s(用一个二进制表示，第i位为1表示当前有第i个钥匙），和当前的x,y坐标,以这三个状态来分层
那么设dist(s,x,y)表示到达当前状态的最短时间
转移就跟BFS的转移一样，除了判断条件里要加关于dist的判断，详见代码
代码
最短路:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<bitset>
#define maxn 15
#define INF 0x3f3f3f3f
using namespace std;
struct node{
int x;
int y;
int k;
node(){
}
node(int xx,int yy,int key){
x=xx;
y=yy;
k=key;
}
};
int n,m,p,k,s;
int dist[2505][maxn][maxn];
int inq[2505][maxn][maxn];
int key[maxn][maxn];
int door[maxn][maxn][maxn][maxn];
const int walkx[4]={1,-1,0,0};
const int walky[4]={0,0,1,-1};
int spfa(){
queue<node>q;
q.push(node(1,1,key[1][1]));
inq[key[1][1]][1][1]=1;
memset(dist,0x3f,sizeof(dist));
dist[key[1][1]][1][1]=0;
while(!q.empty()){
node now=q.front();
q.pop();
inq[now.k][now.x][now.y]=0;
for(int i=0;i<4;i++){
int x=now.x+walkx[i];
int y=now.y+walky[i];
if(x>=1&&x<=n&&y>=1&&y<=m){
if((door[now.x][now.y][x][y]>=1 && (now.k & (1<<door[now.x][now.y][x][y])) !=0)||door[now.x][now.y][x][y]==-1){
if(dist[now.k|key[x][y]][x][y]>dist[now.k][now.x][now.y]+1){
dist[now.k|key[x][y]][x][y]=dist[now.k][now.x][now.y]+1;
if(!inq[now.k|key[x][y]][x][y]){
inq[now.k|key[x][y]][x][y]=1;
q.push(node(x,y,now.k|key[x][y]));
}
}
}
}
}
}
int ans=INF;
for(int i=0;i<(1<<(p+1));i++){
ans=min(ans,dist[i][n][m]);
}
if(ans==INF) return -1;
else return ans;
}
int main(){
int x1,y1,x2,y2,g;
scanf("%d %d %d",&n,&m,&p);
scanf("%d",&k);
memset(door,0xff,sizeof(door));
for(int i=1;i<=k;i++){
scanf("%d %d %d %d %d",&x1,&y1,&x2,&y2,&g);
door[x1][y1][x2][y2]=g;
door[x2][y2][x1][y1]=g;
}
scanf("%d",&s);
for(int i=1;i<=s;i++){
scanf("%d %d %d",&x1,&y1,&g);
key[x1][y1]|=(1<<g);
}
printf("%d
",spfa());
}
状压BFS：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<bitset>
#define maxn 15
using namespace std;
struct node{
int x;
int y;
int k;
int t;
node(){
}
node(int xx,int yy,int key,int tim){
x=xx;
y=yy;
k=key;
t=tim;
}
void debug(){
printf("(%d,%d) tim=%d ",x,y,t);
cout <<"key="<<bitset<10>(k)<<endl;
}
};
int n,m,p,k,s;
int vis[2005][maxn][maxn];
int key[maxn][maxn];
int door[maxn][maxn][maxn][maxn];
const int walkx[4]={1,-1,0,0};
const int walky[4]={0,0,1,-1};
int bfs(){
queue<node>q;
q.push(node(1,1,key[1][1],0));
vis[key[1][1]][1][1]=1;
while(!q.empty()){
node now=q.front();
q.pop();
// now.debug();
if(now.x==n&&now.y==m){
return now.t;
}
for(int i=0;i<4;i++){
int x=now.x+walkx[i];
int y=now.y+walky[i];
if(x>=1&&x<=n&&y>=1&&y<=m){
if(door[now.x][now.y][x][y]==0) continue;
if((door[now.x][now.y][x][y]>=1 && (now.k & (1<<door[now.x][now.y][x][y])) !=0)||door[now.x][now.y][x][y]==-1){
if(vis[now.k|key[x][y]][x][y]) continue;
vis[now.k|key[x][y]][x][y]=1;
q.push(node(x,y,now.k|key[x][y],now.t+1));
}
}
}
}
return -1;
}
int main(){
int x1,y1,x2,y2,g;
scanf("%d %d %d",&n,&m,&p);
scanf("%d",&k);
memset(door,0xff,sizeof(door));
for(int i=1;i<=k;i++){
scanf("%d %d %d %d %d",&x1,&y1,&x2,&y2,&g);
door[x1][y1][x2][y2]=g;
door[x2][y2][x1][y1]=g;
}
scanf("%d",&s);
for(int i=1;i<=s;i++){
scanf("%d %d %d",&x1,&y1,&g);
key[x1][y1]|=(1<<g);
}
printf("%d
",bfs());
}