AOJ 1320 City Merger

有點麻煩的題目，算出 overlap，然後再計算有沒有子字串在裡面，子字串可以直接不用考慮

那麼把剩餘的點拿來做 Hamilton Path 最小成本即可

Copy
/\*
\* GCA : "Computer is artificial subject absolutely,Math is God"
\*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define VAR(a,b) \_\_typeof(b) a=(b)
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#else
#define VAR(a,b) \_\_typeof(b) a=(b)
#define debug(...)
#endif
typedef unsigned int uint;
typedef long long int Int;
typedef unsigned long long int UInt;
#define Set(a,s) memset(a,s,sizeof(a))
#define Pln() printf("\\n")
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define M 25
#define PB push\_back
#define oo INT\_MAX
#define FOR(a,b) for(VAR(a,(b).begin());a!=(b).end();++a)
#define eps 1e-9
#define X first
#define Y second
inline bool xdy(double x,double y){return x>y+eps;}
inline bool xddy(double x,double y){return x>y-eps;}
inline bool xcy(double x,double y){return x<y-eps;}
inline bool xcdy(double x,double y){return x<y+eps;}
const Int mod=1000000007;
int n;
char str\[M\]\[M\];
int mz\[M\]\[M\];
int pnt\[M\];
bool nonused\[M\];
bool substr(int x,int y){
int xlen=strlen(str\[x\]);
int ylen=strlen(str\[y\]);
string sx(str\[x\]),sy(str\[y\]);
if(ylen>xlen)return false;
for(int i=0;i<=xlen-ylen;i++){
if(sx.substr(i,ylen)==sy)return true;
}
return false;
}
int count\_w(int x,int y){
int len=strlen(str\[x\]);
int len2=strlen(str\[y\]);
for(int i=max(0,len-len2);str\[x\]\[i\];i++){
bool suc=true;
int j,k;
for(j=i,k=0;str\[x\]\[j\]&&str\[y\]\[k\];j++,k++){
if(str\[x\]\[j\]!=str\[y\]\[k\]){
suc=false;
break;
}
}
if(suc)return k;
}
return 0;
}
int dp\[1<<15\]\[15\];
int dfs(int s,int now,int dep){
// debug("%d %d\\n",now,dep);
if(s==(1<<n)-1){
return 0;
}
if(dp\[s\]\[now\]!=-1)return dp\[s\]\[now\];
int minans=-1,t;
for(int i=0;i<n;i++){
if(i!=now&&!((s>>i)&1)){
int h;
if((h=dfs(s|(1<<i),i,dep+1))!=-1){
t=h+strlen(str\[pnt\[i\]\])-mz\[now\]\[i\];
}
if(minans==-1||t<minans){
minans=t;
}
}
}
return dp\[s\]\[now\]=minans;
}
void solve(){
Set(nonused,0);
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
if(i==j)continue;
if(substr(i,j))nonused\[j\]=1;
}
}
int cnt=0;
for(int i=0;i<n;i++){
if(!nonused\[i\])pnt\[cnt++\]=i;
}
n=cnt;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
if(i==j)continue;
mz\[i\]\[j\]=count\_w(pnt\[i\],pnt\[j\]);
// debug("%d %d %d\\n",i,j,mz\[i\]\[j\]);
}
}
Set(dp,-1);
int ans=oo;
// ans=dfs(1,0,0);
for(int i=0;i<n;i++){
int t=strlen(str\[pnt\[i\]\])+dfs(1<<i,i,0);
ans=min(ans,t);

}
printf("%d\\n",ans);
}
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%d",&n)&&n){
for(int i=0;i<n;i++)scanf("%s",str\[i\]);
Set(mz,0);
solve();
}

}