[UVALive4394]String Painter
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题目大意:两个字符串,a是原串,b是目标串,一次可以把a的任意段所有字母变成同一个字母,问用最少次数将a变成b。
解题思路:
考虑第j位
如果a[j]!=b[j]
则第j位一定要刷成b[j]
假设是从第i位刷到第j位(i和j要满足b[i]等于b[j])
那么又要求从i+1到j-1位刷成正确的步数
但是这时候a[i+1]到a[j-1]已经变成了b[j]
也就是说一步染色后实质上会对原字符串造成影响
根据上面所得到的
令f[j]表示前j位刷好的最小步数
若a[j]=b[j]则f[j]=f[j-1]
否则找到i满足b[i]=b[j]并且求dp[i+1][j-1][b[j]]
其中dp[i+1][j-1][b[j]]表示在区间[i+1,j-1]被染成b[j]色的情况下达成目标的最小步数
至于dp[][][]怎么求比较简单就不说了
#include<iostream>#include<cstring>#include<cstdio>using namespace std;int dp[110][110][150], f[110];char b[110], a[110];int cal(int i, int j, int c){ if (i > j) return 0; if (dp[i][j][c] != 0x3f3f3f3f) return dp[i][j][c]; if (i == j) { if (c == b[i]) dp[i][j][c] = 0; else dp[i][j][c] = 1; return dp[i][j][c]; } int k; for (k = i; k <= j - 1; k++) dp[i][j][c] = min(dp[i][j][c], cal(i, k, c) + cal(k + 1, j, c)); if (b[i] == b[j]) { if (b[i] == c) dp[i][j][c] = min(dp[i][j][c], dp[i + 1][j - 1][c]); else dp[i][j][c] = min(dp[i][j][c], cal(i + 1, j - 1, b[i]) + 1); } return dp[i][j][c];}int main(){ while (cin >> a) { cin >> b; int len = strlen(b); int i, j; memset(dp, 0x3f, sizeof(dp)); memset(f, 0x3f, sizeof(f)); for (i = 0; i < len; i++) { if (a[i] == b[i]) { if (i == 0) f[i] = 0; else f[i] = f[i - 1]; } else { for (j = i; j >= 0; j--) if (b[i] == b[j]) { if (j == 0) f[i] = min(f[i], 1 + cal(j + 1, i - 1, b[i])); else f[i] = min(f[i], f[j - 1] + 1 + cal(j + 1, i - 1, b[i])); } } } cout << f[len - 1] << endl; } }
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