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;
}

UVA 10453 Make Palindrome(区间简单DP),布布扣,bubuko.com

UVA 10453 Make Palindrome(区间简单DP)

标签：style   blog   class   code   int   string   

原文地址：http://blog.csdn.net/zxjcarrot/article/details/24835935