题意很简单,不说了。
这题思路其实挺简单,把每个格编号为i,3*i走到3*j+1，3*i+1走到3*j+2，以此类推，注意一下边权是多少就好了.
然而一开始用分层图的普遍套路i+j*n*n编号调了好久发现不同层编号有冲突,mdzz....
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <cmath> 6 #include <queue> 7 #include <map> 8 #define ll long long 9 #define out(a) printf("%d ",a) 10 #define ln printf(" ") 11 #define clear(a,b) memset(a,b,sizeof(a)) 12 const int N=1e5+2e4+50; 13 const int MOD=1e9+7; 14 using namespace std; 15 int n,nf,t; 16 int x_,y_,l,r,cnt=0; 17 int x[150][150],a[150][150]; 18 int tot=0; 19 int head[N],dis[N]; 20 bool vis[N]; 21 int ans=23333333; 22 int dx[4]={1,-1,0,0}; 23 int dy[4]={0,0,1,-1}; 24 struct node 25 { 26 int to,nxt,cost; 27 }edge[N<<1]; 28 struct dist 29 { 30 int h,id; 31 bool operator<(const dist&a)const{ 32 return a.h<h; 33 } 34 }; 35 int read() 36 { 37 int s=0,t=1; char c; 38 while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();} 39 while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();} 40 return s*t; 41 } 42 ll readl() 43 { 44 ll s=0,t=1; char c; 45 while (c<'0'||c>'9'){if (c=='-') t=-1; c=getchar();} 46 while (c>='0'&&c<='9'){s=s*10+c-'0'; c=getchar();} 47 return s*t; 48 } 49 void add(int x,int y,int z) 50 { 51 edge[++tot].to=y; 52 edge[tot].cost=z; 53 edge[tot].nxt=head[x]; 54 head[x]=tot; 55 } 56 priority_queue<dist>q; 57 void Dijkstra(int s) 58 { 59 int x,y,z; 60 clear(dis,127); clear(vis,false); 61 dis[s]=0; 62 dist p; 63 p.h=0; p.id=s; 64 q.push(p); 65 while (!q.empty()){ 66 p=q.top(); q.pop(); 67 x=p.id; 68 if (!vis[x]) { 69 vis[x]=true; 70 for (int i=head[x];i;i=edge[i].nxt){ 71 y=edge[i].to; z=edge[i].cost; 72 if (dis[y]>dis[x]+z){ 73 dis[y]=dis[x]+z; 74 p.h=dis[y]; p.id=y; 75 q.push(p); 76 } 77 } 78 } 79 } 80 ans=min(min(dis[3*nf],dis[3*nf+1]),dis[3*nf+2]); 81 } 82 int main() 83 { 84 n=read(),t=read(); 85 for (int i=1;i<=n;i++) 86 for (int j=1;j<=n;j++) 87 x[i][j]=read(),a[i][j]=++cnt; 88 nf=n*n; 89 for (int i=1;i<=n;i++) 90 for (int j=1;j<=n;j++){ 91 for (int k=0;k<4;k++){ 92 x_=i+dx[k]; y_=j+dy[k]; 93 if (x_>0&&x_<=n&&y_>0&&y_<=n) { 94 l=a[i][j]; r=a[x_][y_]; 95 add(3*l,3*r+1,t); //add(3*r+1,3*l,t); 96 add(3*l+1,3*r+2,t); //add(3*r+2,3*l+1,t); 97 add(3*l+2,3*r,x[x_][y_]+t); //add(3*r,3*l+2,x[i][j]+t); 98 } 99 } 100 } 101 Dijkstra(3); 102 out(ans); 103 return 0; 104 } 105 106 107