51Nod 编辑距离 DP+滚动数组

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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;}

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