uva10192 - Vacation(dp,lcs)
2013-06-03 19:38
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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:
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 105 char 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; }
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