Best Sequence
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 5543 Accepted: 2188
Description
The twenty-first century is a biology-technology developing century. One of the most attractive and challenging tasks is on the gene project, especially on gene sorting program. Recently we know that a gene is made of DNA. The nucleotide bases from which DNA is built are A(adenine), C(cytosine), G(guanine), and T(thymine). Given several segments of a gene, you are asked to make a shortest sequence from them. The sequence should use all the segments, and you cannot flip any of the segments.
For example, given ‘TCGG’, ‘GCAG’, ‘CCGC’, ‘GATC’ and ‘ATCG’, you can slide the segments in the following way and get a sequence of length 11. It is the shortest sequence (but may be not the only one).
Input
The first line is an integer T (1 <= T <= 20), which shows the number of the cases. Then T test cases follow. The first line of every test case contains an integer N (1 <= N <= 10), which represents the number of segments. The following N lines express N segments, respectively. Assuming that the length of any segment is between 1 and 20.
Output
For each test case, print a line containing the length of the shortest sequence that can be made from these segments.
Sample Input
1
5
TCGG
GCAG
CCGC
GATC
ATCG
Sample Output
11
之所以说是DP，是因为这道题目实现把每两个字符串连接的状态保存起来，而后进行dfs遍历，最后选出最优值
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
using namespace std;
char a[11][25];
int dp[11][11];
int n;
int ans;
int vis[15];
void add(int m,int n)
{
int k=0;
int len1=strlen(a[m]);
int len2=strlen(a[n]);
bool tag;
for(int p=1;p<=len1&&p<=len2;p++)
{
tag=true;
for(int i=0,j=len1-p;i<p;i++,j++)
{
if(a[m][j]!=a[n][i])
{
tag=false;
break;
}
}
if(tag)
k=p;
}
dp[m][n]=len2-k;
}
void dfs(int pre,int num,int sum)
{
if(num==n)
{
if(ans>sum)
ans=sum;
return;
}
for(int i=0;i<n;i++)
{
if(vis[i]==0)
{
vis[i]=1;
dfs(i,num+1,sum+dp[pre][i]);
vis[i]=0;
}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
getchar();
for(int i=0;i<n;i++)
scanf("%s",a[i]);
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
add(i,j);
memset(vis,0,sizeof(vis));
ans=9999;
for(int i=0;i<n;i++)
{
vis[i]=1;
dfs(i,1,strlen(a[i]));
vis[i]=0;
}
printf("%d
",ans);
}
return 0;
}