具体看题解 我说不清楚 首先可以证明 这个环必然包裹住了所有左上角到其他城市的最短路 证明可以调整为包进去 而不会边劣 然后求出所有最短路 然后把它们包进去的话 把一个交点拆成四个点 然后跑一下左上角绕一圈到左上角的最短路
#include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include<queue> #define cl(x) memset(x,0,sizeof(x)) using namespace std; typedef long long ll; typedef pair<ll,int> abcd; inline char nc(){ static char buf[100000],*p1=buf,*p2=buf; return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++; } inline void read(int &x){ char c=nc(),b=1; for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1; for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b; } const int N=402*402*4+5; const int M=N*4; struct edge{ int u,v,w,next; }G[M]; int head[N],inum; inline void add(int u,int v,int w,int p){ G[p].u=u; G[p].v=v; G[p].w=w; G[p].next=head[u]; head[u]=p; } inline void link(int u,int v,int w){ add(u,v,w,++inum); add(v,u,w,++inum); } #define V G[p].v const int _N=405; int n,m; int a[_N][_N],b[_N][_N],c[_N][_N]; int tot; ll dis[N]; int pre[N]; priority_queue<abcd,vector<abcd>,greater<abcd> > Q; inline void Dij(int S){ for (int i=1;i<=tot;i++) dis[i]=1LL<<60,pre[i]=0; dis[S]=0; Q.push(abcd(0,S)); while (!Q.empty()){ abcd t=Q.top(); Q.pop(); int u=t.second; if (dis[u]!=t.first) continue; for (int p=head[u];p;p=G[p].next) if (dis[V]>dis[u]+G[p].w) pre[V]=u,Q.push(abcd(dis[V]=dis[u]+G[p].w,V)); } } #define P(x,y) ((m+1)*((x)-1)+(y)) #define PP(x,y,z) ((P(x,y)-1)*4+z+1) int vst[N]; int mark[N][4],del[N]; inline void Link(int u,int v,int w){ if (del[u] || del[v]) return; link(u,v,w); } inline void dfs(int u){ if (u==1 || vst[u]) return; vst[u]=1; if (pre[u]==u-1) mark[u][3]=mark[pre[u]][1]=1; if (pre[u]==u+1) mark[u][1]=mark[pre[u]][3]=1; if (pre[u]==u-m-1) mark[u][0]=mark[pre[u]][2]=1; if (pre[u]==u+m+1) mark[u][2]=mark[pre[u]][0]=1; dfs(pre[u]); } int main(){ freopen("t.in","r",stdin); freopen("t.out","w",stdout); read(n); read(m); tot=(n+1)*(m+1); for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) read(a[i][j]); for (int i=1;i<=n;i++) for (int j=1;j<=m+1;j++) read(b[i][j]),link(P(i,j),P(i+1,j),b[i][j]); for (int i=1;i<=n+1;i++) for (int j=1;j<=m;j++) read(c[i][j]),link(P(i,j),P(i,j+1),c[i][j]); Dij(1); for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) if (a[i][j]) dfs(P(i,j)); cl(head); inum=0; for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) if (a[i][j]) del[PP(i,j,2)]=del[PP(i+1,j,1)]=del[PP(i,j+1,3)]=del[PP(i+1,j+1,0)]=1; del[1]=1; for (int i=1;i<=n+1;i++) for (int j=1;j<=m+1;j++) for (int k=0;k<4;k++) if (!mark[P(i,j)][k]) Link(PP(i,j,k),PP(i,j,(k+1)%4),0); for (int i=1;i<=n;i++) for (int j=1;j<=m+1;j++) Link(PP(i,j,3),PP(i+1,j,0),b[i][j]),Link(PP(i,j,2),PP(i+1,j,1),b[i][j]); for (int i=1;i<=n+1;i++) for (int j=1;j<=m;j++) Link(PP(i,j,1),PP(i,j+1,0),c[i][j]),Link(PP(i,j,2),PP(i,j+1,3),c[i][j]); tot<<=2; Dij(2); printf("%lld\n",dis[4]); /*int t=4; while (t) printf("%d\n",t),t=pre[t];*/ return 0; }