题意：
告诉你两个字符串， 问你从第一个字符串刷到第二个字符串最少需要几步？ 每次刷操作： 可以把任意一个区间 刷成同一个字母。
思路：
和LightOJ 1422 神似。点击打开链接
这个只是多了一步。真的感觉区间dp 很神奇。
从一个字符串到另一个字符串 不太好操作。
我们先考虑由空字符串刷到 第二个字符串的最少操作。
先令dp[i][j]表示i~j 区间 空字符串刷到 目标串的最少操作。
那么j 个点，可以单独刷 dp[i][j] = dp[i][j-1] + 1;
也可以由前面和它一样的同时刷。 dp[i][j] = min{dp[i][k] + dp[k+1][j-1]} str2[k] == str2[j]
然后在求字符串1 到目标串 的最少操作。
也是一个递推的过程。
令ans[i] 表示把 起始串 前缀[0,,,,i] 刷到目标串的最少操作。
如果str1[i] == str2[i] ans[i] = ans[i-1]
否则 就在前面一个位置 找到最小值 ans[i] = min{ans[j] + dp[j+1][i]}
注意：
如果是记忆话搜索的话， 很难保证 每个区间 都处理到。
直接预处理 10000个dfs 即可。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100 + 10;
char s[maxn], t[maxn];
int dp[maxn][maxn];
int dfs(int l,int r){
int& ans = dp[l][r];
if (ans != -1) return ans;
if (l > r) return ans = 0;
if (l == r){
return ans = 1;
}
ans = dfs(l,r-1) + 1;
for (int i = l; i < r; ++i){
if (t[i] == t[r]){
ans = min(ans, dfs(i+1, r-1) + dfs(l,i));
}
}
return ans;
}
int ans[maxn];
int main(){
while(~scanf("%s%s",s,t)){
int len = strlen(s);
memset(dp,-1,sizeof dp);
for (int i = 0; i < len; ++i){
for (int j = i; j < len; ++j){
dfs(i,j);
}
}
if (s[0] == t[0]) ans[0] = 0;
else ans[0] = 1;
for (int i = 1; i < len; ++i){
if (s[i] == t[i]){
ans[i] = ans[i-1];
}
else {
ans[i] = dp[0][i];
for (int j = 0; j < i; ++j){
// printf("%d %d %d %d\n", ans[j], dp[j+1][i],j+1,i);
ans[i] = min(ans[i], ans[j] + dp[j+1][i]);
}
}
}
printf("%d\n", ans[len-1]);
}
return 0;
}
String painter
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4251 Accepted Submission(s): 1991
Problem Description
There are two strings A and B with equal length. Both strings are made up of lower case letters. Now you have a powerful string painter. With the help of the painter, you can change a segment of characters of a string to any other character you want. That is, after using the painter, the segment is made up of only one kind of character. Now your task is to change A to B using string painter. What’s the minimum number of operations?
Input
Input contains multiple cases. Each case consists of two lines:
The first line contains string A.
The second line contains string B.
The length of both strings will not be greater than 100.
Output
A single line contains one integer representing the answer.
Sample Input
zzzzzfzzzzz abcdefedcba abababababab cdcdcdcdcdcd
Sample Output
6 7
Source
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