HDU-1503 Advanced Fruits(LCS)

题意：

给两个字符串，让你寻找一个最短的字符串，满足其子序列能够构成这两个字符串。

思路：

简单lcs，求出最长公共子序列之后，将两个串构造在一起即可。

代码：

#include <bits/stdc++.h>using namespace std;char s1[105], s2[105];int len1, len2;string ans, tmp;int dp[105][105];bool col[2][105];void lcs(){memset(dp, 0, sizeof dp);memset(col, 0, sizeof col); for(int i = 1; i <= len1; ++i)for(int j = 1; j <= len2; ++j){if(s1[i] == s2[j])dp[i][j] = dp[i-1][j-1]+1;else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);}int x = len1, y = len2;while(x && y){if(s1[x] == s2[y]){col[0][x] = 1;col[1][y] = 1;--x, --y;}else if(dp[x-1][y] > dp[x][y-1])--x;else--y;}}int main(){while(cin >> s1+1 && cin >> s2+1){len1 = strlen(s1+1);len2 = strlen(s2+1);lcs();ans = "";int x = 1, y = 1;while(1){tmp = "";while(x <= len1 && !col[0][x])tmp += s1[x++];ans += tmp;tmp = "";while(y <= len2 && !col[1][y])tmp += s2[y++];ans += tmp;if(x > len1 && y > len2) break;ans += s1[x];++x, ++y;}cout << ans << endl;}return 0;}

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