| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 8383 | Accepted: 4061 |
Description
Keeping track of all the cows can be a tricky task so Farmer John has installed a system to automate it. He has installed on each cow an electronic ID tag that the system will read as the cows pass by a scanner. Each ID tag's contents are currently a single string with length M (1 ≤ M ≤ 2,000) characters drawn from an alphabet of N (1 ≤ N ≤ 26) different symbols (namely, the lower-case roman alphabet).
Cows, being the mischievous creatures they are, sometimes try to spoof the system by walking backwards. While a cow whose ID is "abcba" would read the same no matter which direction the she walks, a cow with the ID "abcb" can potentially register as two different IDs ("abcb" and "bcba").
FJ would like to change the cows's ID tags so they read the same no matter which direction the cow walks by. For example, "abcb" can be changed by adding "a" at the end to form "abcba" so that the ID is palindromic (reads the same forwards and backwards). Some other ways to change the ID to be palindromic are include adding the three letters "bcb" to the begining to yield the ID "bcbabcb" or removing the letter "a" to yield the ID "bcb". One can add or remove characters at any location in the string yielding a string longer or shorter than the original string.
Unfortunately as the ID tags are electronic, each character insertion or deletion has a cost (0 ≤ cost ≤ 10,000) which varies depending on exactly which character value to be added or deleted. Given the content of a cow's ID tag and the cost of inserting or deleting each of the alphabet's characters, find the minimum cost to change the ID tag so it satisfies FJ's requirements. An empty ID tag is considered to satisfy the requirements of reading the same forward and backward. Only letters with associated costs can be added to a string.
Input
Line 2: This line contains exactly M characters which constitute the initial ID string
Lines 3..N+2: Each line contains three space-separated entities: a character of the input alphabet and two integers which are respectively the cost of adding and deleting that character.
Output
Sample Input
3 4 abcb a 1000 1100 b 350 700 c 200 800
Sample Output
900
Hint
Source
1、dp[i+1][j]表示区间i+1到区间j已经是回文串了的最小代价，那么对于a[i]这个字母，我们有两种操作，删除与添加，对应有两种代价，dp[i+1][j]+add[a[i]],dp[i+1][j]+sub[a[i]]，取这两种代价的最小值；
2、dp[i][j-1]表示区间i到区间j-1已经是回文串了的最小代价，那么对于s[j]这个字母，同样有两种操作，dp[i][j-1]+add[a[j]],dp[i][j-1]+sub[a[j]]，取最小值
3、若是a[i]==a[j]，dp[i+1][j-1]表示区间i+1到区间j-1已经是回文串的最小代价，那么对于这种情况，我们考虑dp[i][j]与dp[i+1][j-1]的大小........
然后dp[i][j]取上面这些情况的最小值
1 /****************************** 2 code by drizzle 3 blog: www.cnblogs.com/hsd-/ 4 ^ ^ ^ ^ 5 O O 6 ******************************/ 7 //#include<bits/stdc++.h> 8 #include<iostream> 9 #include<cstring> 10 #include<cmath> 11 #include<cstdio> 12 #include<bitset> 13 #include<map> 14 //#define ll long long 15 #define mod 1000000007 16 #define PI acos(-1.0) 17 using namespace std; 18 int n,m; 19 char a[2005]; 20 map<char,int> add; 21 map<char,int> sub; 22 char ch; 23 int ad,su; 24 int dp[2005][2005]; 25 int main() 26 { 27 while(scanf("%d %d",&n,&m)!=EOF){ 28 getchar(); 29 add.clear(); 30 sub.clear(); 31 for(int i=1;i<=m;i++) scanf("%c",&a[i]); 32 getchar(); 33 for(int i=1;i<=n;i++) 34 { 35 scanf("%c %d %d",&ch,&ad,&su); 36 add[ch]=ad; 37 sub[ch]=su; 38 getchar(); 39 } 40 int maxn=0; 41 memset(dp,0,sizeof(dp)); 42 for(int i=m;i>=1;i--) 43 { 44 45 for(int j=i+1;j<=m;j++) 46 { 47 48 dp[i][j]=min(dp[i][j-1]+add[a[j]],dp[i][j-1]+sub[a[j]]); 49 dp[i][j]=min(dp[i][j],min(dp[i+1][j]+add[a[i]],dp[i+1][j]+sub[a[i]])); 50 if(a[i]==a[j]) 51 dp[i][j]=min(dp[i][j],dp[i+1][j-1]); 52 } 53 } 54 cout<<dp[1][m]<<endl; 55 } 56 return 0; 57 }