分析

网络流24题总结

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\n",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\n",bfs());
}