hdu2476--string Painter

String Painter Time limit:5000/2000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 1834 Accepted Submission (s): 814


Problem Descriptionthere is strings A and B with equal length. Both strings is 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 of all other character you want. That's, after using the painter, the segment are made up for only one kind of character. Now the your task is to the change A to B using string painter. What ' s the minimum number of operations?
Inputinput contains multiple cases. Each case consists of lines:
The first line contains string A.
The second line contains string B.
The length of both strings would not be greater than 100.

Outputa single line contains one integer representing the answer.
Sample Input

Zzzzzfzzzzzabcdefedcbaababababababcdcdcdcdcdcd

Sample Output

67

Source2008 Asia Regional Chengdu
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Statistic | Submit | Discuss | Note

Interval DP, the problem is very special Ah, first consider from the empty string into a second string, and then use this information to calculate the first string into a second string of the minimum number of times

#include <map> #include <set> #include <list> #include <queue> #include <stack> #include <vector> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring># Include <iostream> #include <algorithm>using namespace std;const int N = 110;const int inf = 0x3f3f3f3f;int dp[ N][n];int F[n];char Stra[n], Strb[n];int main () {while (~scanf ("%s%s", Stra + 1, STRB + 1)) {int N = strlen (stra + 1); memset (DP, 0, sizeof (DP)), memset (f, INF, sizeof (f)), for (int i = 1; I <= n; ++i) {dp[i][i] = 1;} for (int i = n; i >= 1; i.) {for (int j = i + 1; j <= N; ++j) {Dp[i][j] = dp[i + 1][j] + 1;for (int k = i + 1; k < = J; ++k) {if (strb[k] = = Strb[i]) {dp[i][j] = min (Dp[i][j], dp[i + 1][k] + dp[k + 1][j]);}}} F[0] = 0;for (int i = 1; I <= n; ++i) {f[i] = dp[1][i];} for (int i = 1; I <= n; ++i) {if (stra[i] = = Strb[i]) {f[i] = f[i-1];} else{for (int j = 0; J < i; ++j) {F[i] = min (F[i], f[j] + dp[j + 1][i]);}} printf ("%d\n", F[n]);} return 0;}

hdu2476--string Painter