BZOJ 1789 Y形项链(思维)

这题类似于1787，最后的节点一定是两点的LCA，这里也就是两个字符串的最长公共前缀。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=505;
//Code begin...

char s1[N], s2[N], s3[N];

int sol(char* a, char* b, int alen, int blen){
    FO(i,0,min(alen,blen)) if (a[i]!=b[i]) return i;
    return min(alen,blen);
}
int main ()
{
    int len1, len2, len3, ans=INF, tmp, x, y;
    scanf("%d%s%d%s%d%s",&len1,s1,&len2,s2,&len3,s3);
    x=sol(s1,s2,len1,len2); tmp=len1+len2-2*x;
    y=sol(s1,s3,x,len3); tmp+=(len3+x-2*y);
    ans=min(ans,tmp);
    x=sol(s1,s3,len1,len3); tmp=len1+len3-2*x;
    y=sol(s1,s2,x,len2); tmp+=(len2+x-2*y);
    ans=min(ans,tmp);
    x=sol(s2,s3,len2,len3); tmp=len2+len3-2*x;
    y=sol(s1,s2,x,len1); tmp+=(len1+x-2*y);
    ans=min(ans,tmp);
    printf("%d\n",ans);
    return 0;
}