uva10192 - Vacation(dp,lcs)

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最长公共子序列的简单应用。。。

状态：dp[i][j]表示串a[0。。。i]和串b[0。。。j]的最长公共子序列【lcs】。

状态转移：dp[i][j] = {if(a[i]==b[j]) dp[i][j] = dp[i-1][j-1]+1;

if(a[i]!=b[j]) dp[i][j] = max(dp[i-1][j],dp[i][j-1]);}

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define M 105char a[M], b[M];int n1, n2, dp[M][M];int lcm(){    n1 = strlen(a); n2 = strlen(b);    for(int i = 0; i <= n1; i++) dp[i][0] = 0;    for(int i = 0; i <= n2; i++) dp[0][i] = 0;    for(int i = 1; i <= n1; i++)    for(int j = 1; j <= n2; j++)    {        if(a[i-1]==b[j-1]) dp[i][j] = dp[i-1][j-1]+1;        else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);    }    return dp[n1][n2];}int main (){    int cas = 0;    while(gets(a))    {        if(a[0]=='#') break;        gets(b);        printf("Case #%d: you can visit at most %d cities.\n",++cas,lcm());    }    return 0;}