UVaLive 6697 (DP)
2016-08-23 20:46
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题意:给出一个长字符串,再给一个短字符串,进行匹配,如果第i个恰好匹配,则 +8,;如果不匹配,可以给长或短字符串添加-,先后匹配,这样-3,连续的长字符串添加-,需要减去一个4;也可不给添加-,则-5。
思路:dp[i][j][k];k表示0或1,当k==0时,表示长串i位置,短串j位置,敲好匹配; 当k==1时,表示长串i位置,短串j位置,需要添加 - ;的最小值。代码如下
思路:dp[i][j][k];k表示0或1,当k==0时,表示长串i位置,短串j位置,敲好匹配; 当k==1时,表示长串i位置,短串j位置,需要添加 - ;的最小值。代码如下
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <cmath> using namespace std; typedef long long LL; const int maxn = 1007; const int INF = 0x3f3f3f3f; int kase, n; char a[maxn], b[maxn]; int dp[maxn][maxn][2]; int main() { scanf("%d", &kase); while(kase--) { scanf("%s%s", a, b); int lena = strlen(a), lenb = strlen(b); for(int i = 0; i <= lena; ++i) for(int j = 0; j <= lenb; ++j) for(int k = 0; k <= 1; ++k) dp[i][j][k] = -INF; for(int i = 0; i <= lena; ++i) dp[i][0][0] = 0; for(int i = 0; i < lena; ++i) for(int j = 0; j < lenb; ++j) { if(a[i] == b[j]) dp[i+1][j+1][0] = max(dp[i+1][j+1][0], max(dp[i][j][0], dp[i][j][1])+8); else dp[i+1][j+1][0] = max(dp[i+1][j+1][0], max(dp[i][j][0], dp[i][j][1])-5); dp[i][j+1][1] = max(dp[i][j+1][1], dp[i][j][0]-3-4); dp[i][j+1][1] = max(dp[i][j+1][1], dp[i][j][1]-3); dp[i+1][j][1] = max(dp[i+1][j][1], dp[i][j][0]-3-4); dp[i+1][j][1] = max(dp[i+1][j][1], dp[i][j][1]-3); } int ans = 0; for(int i = 0; i <= lena; ++i) ans = max(ans, max(dp[i][lenb][0], dp[i][lenb][1])); printf("%d\n", ans); } }
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