51Nod 编辑距离 DP+滚动数组

51Nod 编辑距离 DP

题目链接：51Nod编辑距离
思路：令dp[i][j] 表示 A字符串前i个字符,与B字符串的前j个字符的最小编辑距离。
那么有，

i = 0 && j == 0 时，
dp[i][j] = 0;
i = 0 && 0 < j < lenB时，
dp[i][j] = j;
 j = 0 && 0 < i < lenA 时，
dp[i][j] = i;
0 < i < lenA && 0 < j < lenB时
if (A[i] == B[j])
dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1]) + 1);
 else
dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;

然后，第i行的结果只与i-1,和i行的结果有关，可以采用一个dp[2][MAXLENB]滚动数组来优化空间。

#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#define FIN             freopen("input.txt","r",stdin)
#define FOUT            freopen("output.txt","w",stdout)

typedef __int64  LL;
//typedef long long LL;
typedef unsigned int uint;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int MAXN = 1000 + 5;
const int MAXM = 1000 + 5;

char A[MAXN], B[MAXN];
int dp[2][MAXN];

int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
while (~scanf("%s %s", A, B)) {
int lenA = strlen(A);
int lenB = strlen(B);
for (int i = 0; i <= lenA; i++) {
for (int j = 0; j <= lenB; j++) {
if (i == 0 && j == 0) dp[i & 1][j] = 0;
else if (i == 0) dp[i & 1][j] = j;
else if (j == 0) dp[i & 1][j] = i;
else if (A[i - 1] == B[j - 1]) dp[i & 1][j] = min(dp[(i - 1) & 1][j - 1], min(dp[(i - 1) & 1][j], dp[i & 1][j - 1]) + 1);
else dp[i & 1][j] = min(dp[(i - 1) & 1][j - 1], min(dp[(i - 1) & 1][j], dp[i & 1][j - 1])) + 1;
}
}
printf("%d\n", dp[lenA & 1][lenB]);
}
return 0;
}