POJ 1795

DNA Laboratory
Time Limit: 5000MS   Memory Limit: 30000K
Total Submissions: 1425   Accepted: 280

Description

Background
Having started to build his own DNA lab just recently, the evil doctor Frankenstein is not quite up to date yet. He wants to extract his DNA, enhance it somewhat and clone himself. He has already figured out how to extract DNA from some of his blood cells, but unfortunately reading off the DNA sequence means breaking the DNA into a number of short pieces and analyzing those first. Frankenstein has not quite understood how to put the pieces together to recover the original sequence.
His pragmatic approach to the problem is to sneak into university and to kidnap a number of smart looking students. Not surprisingly, you are one of them, so you would better come up with a solution pretty fast.
Problem
You are given a list of strings over the alphabet A (for adenine), C (cytosine), G (guanine), and T (thymine),and your task is to find the shortest string (which is typically not listed) that contains all given strings as substrings.
If there are several such strings of shortest length, find the smallest in alphabetical/lexicographical order.

Input

The first line contains the number of scenarios.
For each scenario, the first line contains the number n of strings with 1 <= n <= 15. Then these strings with 1 <= length <= 100 follow, one on each line, and they consist of the letters "A", "C", "G", and "T" only.

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the shortest (and smallest) string as described above. Terminate the output for the scenario with a blank line.

Sample Input

1
2
TGCACA
CAT

Sample Output

Scenario #1:
TGCACAT

Source

TUD Programming Contest 2004, Darmstadt, Germany
 
 
好恶心的状态dp啊！！！！
首先对消除能被其他串包含的串，然后对剩下的串进行建边，建一个有向图，长度是位于一个边指向的反方向所对应的串的后缀与边指向的串的相同的最大前缀长度的负数。
设dp[v][s] 是以v为终点，已经访问过集合s所对应的点的最小值，可建立如下方程
dp[v][s] = min(dp[u ][s | （1 << u)] + dis[v][u])       !((1 << u) & s) = = 1
 
  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 using namespace std;
  6 
  7 #define maxn 105
  8 
  9 #define INF 10000
 10 
 11 int n,ca,len,sum;
 12 char s[20][maxn];
 13 int dp[20][(1 << 15) + 5],dis[20][20];
 14 bool vis[20],done[20];
 15 string ans;
 16 
 17 int cal(int x,int y) {
 18         int  _max = 0;
 19         for(int i = 0; i < strlen(s[x]); i++) {
 20                 if(s[x][i] != s[y][0]) continue;
 21                 int j,k;
 22                 for( j = i,k = 0; j < strlen(s[x]) && k < strlen(s[y]); j++,k++) {
 23                         if(s[x][j] != s[y][k]) break;
 24                 }
 25                 if(k == strlen(s[y])) {
 26                         done[y] = 1;
 27                         break;
 28                 }
 29                 if(j == strlen(s[x])) {
 30                         _max = max(_max,j - i);
 31 
 32                 }
 33         }
 34 
 35 
 36         return -_max;
 37 }
 38 void init() {
 39     for(int u = 0; u < n; u++) {
 40             if(done[u]) continue;
 41             for(int v = 0; v < n; v++) {
 42                     if(u == v || done[v]) continue;
 43                     dis[u][v] = cal(u,v);
 44 
 45 
 46             }
 47     }
 48 
 49 
 50 }
 51 
 52 void dfs(int v,int s1) {
 53         vis[v] = 1;
 54         int id = -1;
 55         string t("z");
 56         for(int u = 0; u < n; u++) {
 57                 if(done[u] || vis[u]) continue;
 58 
 59                 if(dp[v][s1] == dp[u][s1 | (1 << u)] + dis[v][u]) {
 60                         string t1(s[u] - dis[v][u],s[u] + strlen(s[u]));
 61                         if(t > t1) {
 62                                 t = t1;
 63                                 id = u;
 64                         }
 65                 }
 66 
 67 
 68         }
 69 
 70         if(id != -1) {
 71                 ans = ans + t;
 72                 dfs(id,s1 | (1 << id));
 73 
 74         }
 75 }
 76 
 77 
 78 
 79 
 80 
 81 void solve() {
 82     init();
 83 
 84     for(int s1 = (1 << n) - 2; s1; s1--) {
 85             for(int v = 0; v < n; v++) {
 86                     if(!(s1 & (1 << v)) || done[v]) continue;
 87                     for(int u = 0; u < n; u++) {
 88                             if(u == v || (s1 & (1 << u)) || done[v] ) continue;
 89                             dp[v][s1] = min(dp[v][s1],dp[u][s1 | (1 << u)] + dis[v][u]);
 90 
 91                     }
 92             }
 93     }
 94 
 95     int _min = 0;
 96     for(int i = 0; i < n; i++) {
 97             if(done[i]) continue;
 98             _min = min(_min,dp[i][1 << i]);
 99     }
100 
101     memset(vis,0,sizeof(vis));
102 
103     ans = "z";
104     int id;
105     for(int i = 0; i < n; i++) {
106             if(done[i]) continue;
107             string t(s[i]);
108             if(dp[i][1 << i] == _min && ans > t) {
109                     ans = t;
110                     id = i;
111             }
112     }
113 
114     dfs(id,1 << id);
115 
116     printf("Scenario #%d:\n",ca++);
117     cout << ans << endl;
118 
119 
120 
121 }
122 int main()
123 {
124     int t;
125     //freopen("sw.in","r",stdin);
126     scanf("%d",&t);
127     ca = 1;
128 
129     while(t--) {
130             memset(done,0,sizeof(done));
131 
132             scanf("%d",&n);
133 
134             for(int i = 0; i < n; i++) {
135                     scanf("%s",s[i]);
136             }
137 
138             memset(dis,0,sizeof(dis));
139 
140             for(int i = 0; i < n; i++) {
141                     for(int s = 0; s < (1 << n); s++) {
142                             dp[i][s] = 0;
143                     }
144             }
145 
146             solve();
147             printf("\n");
148 
149     }
150 
151     return 0;
152 }
View Code

 

posted @ 2014-03-09 13:32  hyx1  阅读(586)  评论(0)  编辑  收藏  举报