[最短路 杂题] BZOJ 4356 Ceoi2014 Wall

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
,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
; int pre
;
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
;
int mark
[4],del
;
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;
}