这个题的搜索可以打到48分……
#include <cstdio>
#include <cstring>
#include <algorithm>
const int N=12;
bool must[N],in[N];
int cnt;
int n,a[N][N],q[N],b[N];
inline bool judge(int len,int lim){
return lim-len>=cnt;
}
inline bool check(int len){
register int i,j,k,pos,g;
register bool can;
for(i=1;i<=n;++i){
can=false;
for(j=1;j<=len;++j){
if(len-j+1>=a[i][0]){
pos=0,g=0;
for(k=j;k<=len;++k){
if(in[q[k]])continue;
if(q[k]!=a[i][pos+1]){g=-1;break;}
b[++pos]=q[k];
in[q[k]]=true;
}
can=g!=-1&&pos==a[i][0];
while(pos--)in[b[pos+1]]=false;
if(can)break;
}
if(j>=a[i][0]){
pos=0,g=0;
for(k=j;k>0;--k){
if(in[q[k]])continue;
if(q[k]!=a[i][pos+1]){g=-1;break;}
b[++pos]=q[k];
in[q[k]]=true;
}
can=g!=-1&&pos==a[i][0];
while(pos--)in[b[pos+1]]=false;
if(can)break;
}
}
if(!can)return false;
}
return true;
}
inline bool dfs(int pos,int lim){
if(!judge(pos-1,lim))return false;
if(pos==lim+1)return check(pos-1);
register int i;
register bool keep;
for(i=1;i<=9;++i)
if(i!=q[pos-1]){
q[pos]=i;
if(must[i]){
must[i]=false,--cnt;
keep=true;
}else keep=false;
if(dfs(pos+1,lim))return true;
if(keep)must[i]=true,++cnt;
}
return false;
}
int main(){
//freopen("rio.in","r",stdin);
scanf("%d",&n);
register int i,j;
for(i=1;i<=n;++i){
while(true){
scanf("%d",&j);
if(!j)break;
a[i][++a[i][0]]=j;
must[j]=true;
}
}
for(i=1;i<=9;++i)
cnt+=must[i];
for(i=cnt;i<=10;++i){
if(cnt<5&&i==10)break;
if(dfs(1,i)){
printf("%d\n",i);
return 0;
}
}
puts("-1");
return 0;
}
Kod
搜索的时候减枝实在是太重要了,其次梦想与搜索之间的平衡也是很重要的,但是一定要合理且适当.
正解很神.
贴一发题解:
![](https://img.laitimes.com/img/9ZDMuAjOiMmIsIjOiQnIsIyZuBnLxgzNxMjMxADMx0SN5UjN1YTN3EjMyMDM4EDMy0CN3gDM4ETMvw1MwgTMwIzLcRzN4ADOxEzLcd2bsJ2Lc12bj5ycn9Gbi52YugTMwIzcldWYtl2Lc9CX6MHc0RHaiojIsJye.png)
还有一位大佬的代码注释也很好:https://loj.ac/submission/66283
我就是看的这两个资料,然后码出来的.(把写法从spfa改成记忆化dfs会快3倍……)
在这里说一下坑吧,就是正着走的状态可以在半截出现,倒着走的状态可以在半截消失.
分享一下代码.
正着走的单向40分:
#include <cstdio>
#include <cstring>
#include <algorithm>
const int N=12;
const int F=2010;
const int Inf=0x3f3f3f3f;
int n,ans,full,bit[N],a[N][N],l[N],b[N];
int f[N][N][F];
bool ex[N][N],can[N][N][N],in[N],fin[N][N][F];
inline bool check(int id,int len,int to){
register int i,pos=0;
memset(in,0,sizeof(in));
for(i=1;i<=len;++i)
in[a[id][i]]=true;
for(i=1;i<=l[to];++i){
if(in[a[to][i]])continue;
b[++pos]=a[to][i];
in[a[to][i]]=true;
}
if(pos!=l[id]-len)return false;
for(i=len+1;i<=l[id];++i)
if(a[id][i]!=b[i-len])
return false;
return true;
}
inline void dfs(int id,int pos,int st){
//printf("%d %d %d\n",id,pos,st);
if(fin[id][pos][st]){
//puts("GG");
ans=std::min(ans,f[id][pos][st]);
return;
}
register int i;
for(i=1;i<=n;++i)
if(i!=id&&(bit[i]&st)==0)
if(can[id][pos][i]&&f[i][0][st|bit[id]]>f[id][pos][st]){
f[i][0][st|bit[id]]=f[id][pos][st];
dfs(i,0,st|bit[id]);
}
if(pos!=l[id]&&f[id][pos+1][st]>f[id][pos][st]+1){
f[id][pos+1][st]=f[id][pos][st]+1;
dfs(id,pos+1,st);
}
}
int main(){
//freopen("rio.in","r",stdin);
scanf("%d",&n);
full=(1<<n)-1;
register int i,j,k;
for(i=1;i<=n;++i){
bit[i]=1<<(i-1);
while(true){
scanf("%d",&j);
if(!j)break;
a[i][++l[i]]=j;
ex[i][j]=true;
}
}
for(i=1;i<=n;++i)
for(j=0;j<=l[i];++j)
for(k=1;k<=n;++k)
can[i][j][k]=check(i,j,k);
memset(f,0x3f,sizeof(f));
ans=Inf;
for(i=1;i<=n;++i)
fin[i][l[i]][full^bit[i]]=true;
for(i=1;i<=n;++i){
f[i][0][0]=0;
dfs(i,0,0);
}
printf("%d\n",ans==Inf?-1:ans);
return 0;
}
Kod
倒着走的单向40分:
#include <cstdio>
#include <cstring>
#include <algorithm>
const int N=12;
const int F=2010;
const int Inf=0x3f3f3f3f;
int n,ans,full,bit[N],a[N][N],l[N],b[N];
int f[N][N][F];
bool can[N][N][N],in[N],fin[N][N][F];
inline bool check(int id,int len,int to){
register int i,pos=0;
memset(in,0,sizeof(in));
for(i=1;i<=len;++i)
in[a[id][i]]=true;
for(i=1;i<=l[to];++i){
if(in[a[to][i]])continue;
b[++pos]=a[to][i];
in[a[to][i]]=true;
}
if(pos!=l[id]-len)return false;
for(i=len+1;i<=l[id];++i)
if(a[id][i]!=b[i-len])
return false;
return true;
}
inline void dfs(int id,int pos,int st){
//printf("%d %d %d\n",id,pos,st);
if(fin[id][pos][st]){
//puts("GG");
ans=std::min(ans,f[id][pos][st]);
return;
}
if(!pos){
register int i,j;
for(i=1;i<=n;++i)
if(i!=id&&(bit[i]&st)==0)
for(j=0;j<=l[i];++j)
if(can[i][j][id]&&f[i][j][st|bit[id]]>f[id][pos][st]){
f[i][j][st|bit[id]]=f[id][pos][st];
dfs(i,j,st|bit[id]);
}
}
if(pos&&f[id][pos-1][st]>f[id][pos][st]+1){
f[id][pos-1][st]=f[id][pos][st]+1;
dfs(id,pos-1,st);
}
}
int main(){
//freopen("rio.in","r",stdin);
scanf("%d",&n);
full=(1<<n)-1;
register int i,j,k;
for(i=1;i<=n;++i){
bit[i]=1<<(i-1);
while(true){
scanf("%d",&j);
if(!j)break;
a[i][++l[i]]=j;
}
}
for(i=1;i<=n;++i)
for(j=0;j<=l[i];++j)
for(k=1;k<=n;++k)
can[i][j][k]=check(i,j,k);
memset(f,0x3f,sizeof(f));
ans=Inf;
for(i=1;i<=n;++i)
fin[i][0][full^bit[i]]=true;
for(i=1;i<=n;++i){
f[i][l[i]][0]=0;
dfs(i,l[i],0);
}
printf("%d\n",ans==Inf?-1:ans);
return 0;
}
Kod
经过我不断修改的清真美丽100分代码:
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long LL;
std::set<LL> ext;
const int N=11;
const int F=1050;
const int Inf=0x3f3f3f3f;
int n,full,bit[N],a[N][N],l[N];
int f[N][N][N][N][F];
bool ex[N][N][N],can[N][N][N],in[N],vis[N][N][N][N][F];
inline bool check(int id,int len,int to){
int i,pos=0;
memcpy(in,ex[id][len],sizeof(in));
for(i=1;i<=l[to];++i){
if(in[a[to][i]])continue;
++pos,in[a[to][i]]=true;
if(a[to][i]!=a[id][len+pos])
return false;
}
if(pos!=l[id]-len)return false;
return true;
}
inline void Init(){
scanf("%d",&n);
int i,j,k;
LL key;
for(i=1;i<=n;++i){
l[i]=0,key=0;
memset(ex[i],0,sizeof(ex[i]));
bit[i]=1<<(i-1),j=1;
while(j){
scanf("%d",&j);
key=key*10+j;
a[i][++l[i]]=j;
for(k=1;k<=l[i];++k)
if(a[i][k])
ex[i][l[i]][a[i][k]]=true;
}
--l[i];
if(ext.count(key))--i,--n;
ext.insert(key);
}
full=(1<<n)-1;
for(i=1;i<=n;++i){
can[0][0][i]=true;
for(j=0;j<=l[i];++j)
for(k=1;k<=n;++k)
can[i][j][k]=check(i,j,k);
}
memset(f,0x3f,sizeof(f));
}
#define cover(id1,pos1,id2,pos2,st,add) (k=std::min(dfs(id1,pos1,id2,pos2,st)+add,k))
inline int dfs(int id1,int pos1,int id2,int pos2,int st){
if(st==full&&pos1==l[id1]&&!pos2)return 0;
int i,j,&k=f[id1][pos1][id2][pos2][st];
if(vis[id1][pos1][id2][pos2][st])return k;
vis[id1][pos1][id2][pos2][st]=true;
if(id2&&!pos2){
for(i=0;i<=n;++i)
if(i!=id1&&!(bit[i]&st))
for(j=0;j<=l[i];++j)
if(can[i][j][id2])
cover(id1,pos1,i,j,st|bit[i],0);
}
for(i=1;i<=n;++i)
if(can[id1][pos1][i]&&!(bit[i]&st))
cover(i,0,id2,pos2,st|bit[i],0);
if(pos1==l[id1]&&!pos2)return k;
if(pos1!=l[id1]&&(ex[id2][pos2+1][a[id1][pos1+1]]||!id2))
cover(id1,pos1+1,id2,pos2,st,1);
if(pos2&&(ex[id1][pos1][a[id2][pos2]]||!id1))
cover(id1,pos1,id2,pos2-1,st,1);
if(pos1!=l[id1]&&pos2&&a[id2][pos2]==a[id1][pos1+1])
cover(id1,pos1+1,id2,pos2-1,st,1);
return k;
}
inline void Work(){
int i,ans=Inf;
for(i=1;i<=n;++i)
ans=std::min(dfs(0,0,i,l[i],bit[i]),ans);
printf("%d\n",ans==Inf?-1:ans);
}
int main(){
Init(),Work();
return 0;
}
Kod
我第一次A掉的时候写的100分(这个代码又臭又长又屎):
#include <cstdio>
#include <cstring>
#include <algorithm>
const int N=12;
const int F=1112;
const int Inf=0x3f3f3f3f;
int n,ans,full,bit[N],a[N][N],l[N],b[N];
int f[N][N][N][N][F];
bool ex[N][N][N],can[N][N][N],in[N],fin[N][N][N][N][F];
inline bool check(int id,int len,int to){
register int i,pos=0;
memset(in,0,sizeof(in));
for(i=1;i<=len;++i)
in[a[id][i]]=true;
for(i=1;i<=l[to];++i){
if(in[a[to][i]])continue;
b[++pos]=a[to][i];
in[a[to][i]]=true;
}
if(pos!=l[id]-len)return false;
for(i=len+1;i<=l[id];++i)
if(a[id][i]!=b[i-len])
return false;
return true;
}
inline void dfs(int id1,int pos1,int id2,int pos2,int st){
//printf("%d %d %d %d %d\n",id1,pos1,id2,pos2,st);
if(fin[id1][pos1][id2][pos2][st]){
//puts("GG");
ans=std::min(ans,f[id1][pos1][id2][pos2][st]);
return;
}
if(id2&&!pos2){
register int i,j;
for(i=1;i<=n;++i)
if(i!=id1&&i!=id2&&(bit[i]&st)==0)
for(j=0;j<=l[i];++j)
if(can[i][j][id2]&&f[id1][pos1][i][j][st|bit[id2]]>f[id1][pos1][id2][pos2][st]){
f[id1][pos1][i][j][st|bit[id2]]=f[id1][pos1][id2][pos2][st];
dfs(id1,pos1,i,j,st|bit[id2]);
}
if(id1){
if(f[id1][pos1][0][0][st|bit[id2]]>f[id1][pos1][id2][pos2][st]){
f[id1][pos1][0][0][st|bit[id2]]=f[id1][pos1][id2][pos2][st];
dfs(id1,pos1,0,0,st|bit[id2]);
}
}else{
for(i=1;i<=n;++i)
if(i!=id2&&(bit[i]&st)==0&&f[i][0][0][0][st|bit[id2]]>f[id1][pos1][id2][pos2][st]){
f[i][0][0][0][st|bit[id2]]=f[id1][pos1][id2][pos2][st];
dfs(i,0,0,0,st|bit[id2]);
}
}
}
if(id1){
register int i;
for(i=1;i<=n;++i)
if(i!=id1&&i!=id2&&(bit[i]&st)==0)
if(can[id1][pos1][i]&&f[i][0][id2][pos2][st|bit[id1]]>f[id1][pos1][id2][pos2][st]){
f[i][0][id2][pos2][st|bit[id1]]=f[id1][pos1][id2][pos2][st];
dfs(i,0,id2,pos2,st|bit[id1]);
}
}else{
register int i;
for(i=1;i<=n;++i)
if(i!=id2&&(bit[i]&st)==0&&f[i][0][id2][pos2][st]>f[id1][pos1][id2][pos2][st]){
f[i][0][id2][pos2][st]=f[id1][pos1][id2][pos2][st];
dfs(i,0,id2,pos2,st);
}
}
if((id2==0||pos2==0)&&(id1==0||pos1==l[id1]))return;
if(id1&&pos1!=l[id1]&&(ex[id2][pos2+1][a[id1][pos1+1]]||!id2)&&f[id1][pos1+1][id2][pos2][st]>f[id1][pos1][id2][pos2][st]+1){
f[id1][pos1+1][id2][pos2][st]=f[id1][pos1][id2][pos2][st]+1;
dfs(id1,pos1+1,id2,pos2,st);
}
if(id2&&pos2&&(ex[id1][pos1+1][a[id2][pos2]]||!id1)&&f[id1][pos1][id2][pos2-1][st]>f[id1][pos1][id2][pos2][st]+1){
f[id1][pos1][id2][pos2-1][st]=f[id1][pos1][id2][pos2][st]+1;
dfs(id1,pos1,id2,pos2-1,st);
}
//printf("<>%d %d %d %d %d\n",id1,pos1,id2,pos2,st);
if(id1&&id2&&pos2&&pos1!=l[id1]&&a[id2][pos2]==a[id1][pos1+1]&&f[id1][pos1+1][id2][pos2-1][st]>f[id1][pos1][id2][pos2][st]+1){
//puts("OK");
f[id1][pos1+1][id2][pos2-1][st]=f[id1][pos1][id2][pos2][st]+1;
dfs(id1,pos1+1,id2,pos2-1,st);
}
//else puts("NO");
}
int main(){
//freopen("rio.in","r",stdin);
//freopen("wq.out","w",stdout);
scanf("%d",&n),full=(1<<n)-1;
register int i,j,k;
for(i=1;i<=n;++i){
bit[i]=1<<(i-1);
while(true){
scanf("%d",&j);
if(!j)break;
a[i][++l[i]]=j;
}
}
for(i=1;i<=n;++i)
for(j=0;j<=l[i];++j)
for(k=1;k<=n;++k)
can[i][j][k]=check(i,j,k);
for(i=1;i<=n;++i)
for(j=0;j<=11;++j)
for(k=1;k<=j;++k)
ex[i][j][a[i][k]]=true;
memset(f,0x3f,sizeof(f));
ans=Inf;
for(i=1;i<=n;++i){
for(j=1;j<=n;++j){
if(i==j)continue;
fin[i][l[i]][j][0][full^bit[i]^bit[j]]=true;
f[i][0][j][l[j]][0]=0;
}
fin[i][l[i]][0][0][full^bit[i]]=true;
f[i][0][0][0][0]=0;
fin[0][0][i][0][full^bit[i]]=true;
f[0][0][i][l[i]][0]=0;
}
for(i=1;i<=n;++i){
for(j=1;j<=n;++j){
if(i==j)continue;
dfs(i,0,j,l[j],0);
}
dfs(i,0,0,0,0);
dfs(0,0,i,l[i],0);
}
printf("%d\n",ans==Inf?-1:ans);
return 0;
}
Kod
我在这里说一下我对于这道题的解法的理解吧.
看到题解之后,就会发现这道题dp的状态定义很谜.我们先看正着走单向的,我们之所以这么定义,是因为最终方案也一定是这样构成的,这很好理解.再看倒着走的,同样,我们之所以这么定义,也是因为最终方案也一定是这样构成的.那么我们再看双向的状态,思考一下,发现也是这样的.那么我们这么转移就一定没有问题.而这神奇的状态设计从哪里来呢.我觉得,就是抓出题目给出的操作的性质——“远观不如近看”,让“近看”去满足“远观”,用“远观”去匹配“近看”,再加上增量构造的思路,就迸发出了这道题的状态以及转移.
现在以我的水平,我对于这道题的理解大概也就这么多,也许以后会有更好的理解把.利用状态压缩,以及对于特殊的问题,设计合理的状态,大概就是我从这道题里学到的宝贵的东西,对于设计状态,我还是摸不着头脑的.
转载于:https://www.cnblogs.com/TSHugh/p/8625433.html