Best Sequence

poj1699:http://poj.org/problem?id=1699

题意:给你nge串,让你求出这些串组成的最小的串重叠部分只算一次。

题解:我的做法是DFS,因为数据范围只有10,就算是n!也只有300多万,加上剪枝,就可以过了。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 int n;
 7 char str[100][100],s[10000];
 8 int ans;
 9 bool vis[100];
10 int getlen(char s1[],char s2[],int l1,int l2){
11       int ans;//表示重叠部分的长度
12       if(l1>=l2){
13          for(ans=l2;ans>=0;ans--){
14              bool flag=true;
15            for(int i=0;i<ans;i++){
16                 if(s2[i]!=s1[l1-(ans-i)]){
17                     flag=false;
18                     break;
19                 }
20           }
21           if(flag)return ans;
22       }
23     }
24     else{
25      for(ans=l1;ans>=0;ans--){
26              bool flag=true;
27            for(int i=0;i<ans;i++){
28                 if(s2[i]!=s1[l1-(ans-i)]){
29                     flag=false;
30                     break;
31                 }
32
33           }
34           if(flag)return ans;
35       }
36     }
37 }
38 void DFS(char fa[],int ll,int dep){
39       if(ll>=ans)return;
40       if(dep==n){
41         ans=min(ll,ans);
42         return;
43       }
44       for(int i=1;i<=n;i++){
45           if(vis[i])continue;
46           int tt=strlen(str[i]);
47           int temp=getlen(fa,str[i],ll,tt);
48           for(int j=ll;j<ll+(tt-temp);j++){
49             fa[j]=str[i][temp+(j-ll)];
50           }
51           vis[i]=1;
52           DFS(fa,ll+tt-temp,dep+1);
53           vis[i]=0;
54       }
55 }
56 int main(){
57   int T;
58   scanf("%d",&T);
59   while(T--){
60     scanf("%d",&n);
61     memset(str,0,sizeof(str));
62     memset(s,0,sizeof(s));
63     memset(vis,0,sizeof(vis));
64     ans=10000000;
65     for(int i=1;i<=n;i++){
66         scanf("%s",str[i]);
67     }
68     for(int i=1;i<=n;i++){
69       memset(s,0,sizeof(s));
70        strcpy(s,str[i]);
71         vis[i]=1;
72         DFS(s,strlen(str[i]),1);
73         vis[i]=0;
74     }
75     printf("%d\n",ans);
76   }
77
78 }

时间: 2024-10-23 04:57:29

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