UVA 10453 Make Palindrome(区间简单DP)

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题意:给出一个字符串A,求出需要至少插入多少个字符使得这个字符串变成回文串.

思路:设dp[i][j]为使区间[i, j]变成回文串所需要的最少字符个数.

1.A[i] == A[j的情况]那么dp[i][j] = min(dp[i][j], dp[i + 1][j -1]);

2.或者在第j个位置插入一个字符A[i], dp[i][j] = min(dp[i][j], dp[i][j - 1] + 1);

3.或者在第i个位置插入一个字符A[j], dp[i][j] = min(dp[i][j], dp[i + 1][j] + 1);

最后记录一下路径输出.

#include <cstdio>#include <string.h>#include <algorithm>using namespace std;typedef char byte;const int MAX = 1005;const int INF = 0x20202020;int dp[MAX][MAX];byte path[MAX][MAX];char A[MAX];int dfs(int i, int j){if(i >= j)return 0;else if(dp[i][j] != INF)return dp[i][j];int & ref = dp[i][j];if(A[i] == A[j]){ref = dfs(i + 1, j - 1);path[i][j] = 0;}int t = dfs(i + 1, j) + 1;if(ref > t){ref = t;path[i][j] = 2;}t = dfs(i, j - 1) + 1;if(ref > t){ref = t;path[i][j] = 1;}return ref;}void print_path(int i, int j){if(i > j)return;else if(i == j)printf("%c", A[i]);else if(path[i][j] == 0){printf("%c", A[i]);print_path(i + 1, j - 1);printf("%c", A[j]);}else if(path[i][j] == 1){printf("%c", A[j]);print_path(i, j - 1);printf("%c", A[j]);}else if(path[i][j] == 2){printf("%c", A[i]);print_path(i + 1, j);printf("%c", A[i]);}}int main(int argc, char const *argv[]){while(~scanf("%s", A)){int n = strlen(A);memset(dp, 0x20, sizeof(dp));printf("%d ", dfs(0, n - 1));print_path(0, n - 1);printf("\n");}return 0;}

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