今天做ural 1002，題目描述：
1002. Phone numbers
Time Limit: 2.0 second
Memory Limit: 16 MB
Background
In the present world you frequently meet a lot of call numbers and they are going to be longer and longer. You need to remember such a kind of numbers. One method to do it in an easy way is to assign letters to digits as shown in the following picture:
1 ij 2 abc 3 def
4 gh 5 kl 6 mn
7 prs 8 tuv 9 wxy
0 oqz
This way every word or a group of words can be assigned a unique number, so you can remember words instead of call numbers. It is evident that it has its own charm if it is possible to find some simple relationship between the word and the person itself. So you can learn that the call number 941837296 of a chess playing friend of yours can be read as WHITEPAWN, and the call number 2855304 of your favourite teacher is read BULLDOG.
Problem
Write a program to find the shortest sequence of words (i.e. one having the smallest possible number of words) which corresponds to a given number and a given list of words. The correspondence is described by the picture above.
Input
Input contains a series of tests. The first line of each test contains the call number, the transcription of which you have to find. The number consists of at most 100 digits. The second line contains the total number of the words in the dictionary (maximum is 50000). Each of the remaining lines contains one word, which consists of maximally 50 small letters of the English alphabet. The total size of the input file doesn't exceed 300KB. The last line of input file contains call number -1.
Output
Each line of output contains the shortest sequence of words which has been found by your program. The words are separated by single spaces. If there is no solution to the input data, the line contains text "No solution.". If there are more solutions having the minimum number of words, you can choose any single one of them.
Sample
|
input
|
output
|
7325189087
5
it
your
reality
real
our
4294967296
5
it
your
reality
real
our
-1
|
reality our
No solution.
|
思路就是dp,狀態轉移是dp[i]=min(dp[i-len[j]]+1)(1<=j<=n)
code:
#include <iostream>
#include <string>
#include <algorithm>
#include <stack>
#include <vector>
using namespace std;
//#pragma comment(linker, "/STACK:16777216")
const int N=105;
const int M=50001;
int dp[N];
int pathone[N];
int pathtwo[N];
int hash[26]={2,2,2,3,3,3,4,4,1,1,5,5,6,6,0,7,0,7,7,8,8,8,9,9,9,0};
vector<string> vec,v;
vector<int> ve;
void Init()
{
memset(dp,0,sizeof(dp));
memset(pathone,0,sizeof(pathone));
memset(pathtwo,0,sizeof(pathtwo));
vec.clear();
v.clear();
ve.clear();
};
int main()
{
string ss;
while(cin>>ss){
if(ss=="-1")break;
Init();
int n;
cin>>n;
string str;
for(int i=0;i<n;i++){
cin>>str;
v.push_back(str);
string tmp;
for(int j=0;j<str.size();j++){
tmp+=hash[str[j]-'a']+48;
}
vec.push_back(tmp);
}
dp[0]=1;
for(int i=1;i<=ss.size();++i){
for(int j=0;j<vec.size();++j){
if(i-vec[j].size()>=0&&dp[i-vec[j].size()]>0){
if(vec[j]==ss.substr(i-vec[j].size(),vec[j].size())&&(dp[i]==0||dp[i]>dp[i-vec[j].size()]+1)){
dp[i]=dp[i-vec[j].size()]+1;
pathone[i]=j;
pathtwo[i]=i-vec[j].size();
}
}
}
}
if(dp[ss.size()]==0){
cout<<"No solution.\n";
}
else{
int mm=ss.size();
//ve.push_back(mm);
while(mm!=0){
//stk.push(v[pathone[mm]]);
ve.push_back(mm);
mm=pathtwo[mm];
}
for(int i=ve.size()-1;i>=0;i--){
if(i==ve.size()-1)cout<<v[pathone[ve[i]]];
else cout<<" "<<v[pathone[ve[i]]];
}
cout<<endl;
}
}
return 0;
}
以上code ,(Crash), 看了FAQ,相當于
STACK_OVERFLOW ,先是郁悶。。。。
玩了花了N多時間查找哪錯問題了。。。。
沒辦法時加了句 i-vec[j].size()<N于是奇跡出現了Accepted.....汗
于是想問下各位路過大牛，，題目明顯是i<N,i-vec[j].size()<N不是就成了句廢話了。。。。無語.....
我碰到類似dp,用dp[i+len[j]]>dp[i]+1 dp[i+len[j]]=dp[i]+1轉換下，就不會出現此問題，，而用dp[i-len[j]]+1<dp[i],dp[i]=dp[i-len[j]]+1問題就出現了。。。RE，所以想問下原因，，，我以保證i-len[j]>=0&&i-len[j]<N(dp[N])
posted on 2008-04-05 16:21
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