06 2019 档案
摘要："数的划分" Code： c++ include include include using namespace std; //Mystery_Sky // define M 500 define INF 0x3f3f3f3f int f[M][M], n, k; int main() { scan
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摘要："花店橱窗布置" 里程碑式的一题！ 第一次完全未看题解做出的一道dp题 Code: c++ include include include using namespace std; //Mystery_Sky // define M 500 define INF 0x3f3f3f3f int f[M
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摘要："书的复制" Code： c++ include include include using namespace std; //Mystery_Sky // define M 700 define INF 0x3f3f3f3f int f[M][M], sum[M], a[M]; int n, k;
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摘要："计算字符串距离" 同样也是字符串距离计算问题，参考 "一本通 1276：【例9.20】编辑距离" Code: c++ include include include using namespace std; //Mystery_Sky // define INF 0x3f3f3f3f define
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摘要："滑雪" 记忆化搜索 Code: c++ include include include using namespace std; //Mystery_Sky // define M 500 define INF 0x3f3f3f3f int f[M][M]; int r, c, map[M][M]
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摘要："编辑距离" Code: c++ include include include using namespace std; //Mystery_Sky // define INF 0x3f3f3f3f define M 3000 int f[M][M]; int len_a, len_b; char
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摘要："乘积最大（数据弱化版）" ps:本题无需使用大整数。 Code: c++ include include include using namespace std; //Mystery_Sky // define INF 0x3f3f3f3f define M 500 int f_max[M][M]
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摘要："合并石子" 状态转移方程：f_min[i][i] = 0, f_min[i][j] = min(f_min[i][k] + f_min[k+1][j] + sum[j] sum[i 1]; Code: c++ include include include using namespace std;
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摘要："城市交通路网" 最短路 + 路径输出 Code: c++ include include include include using namespace std; //Mystery_Sky // define M 1000100 define INF 0x3f3f3f3f struct Edge
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摘要："友好城市" 对北岸（或南岸）的城市从小到大排序，再求南岸（或北岸）的城市位置的最长不下降序列长度即可。 ps:这里数据较弱，用n n的做法可以过。 Code: c++ include include include include using namespace std; //Mystery_Sk
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摘要："购买巧克力Chocolate Buying" 乍一看以为是背包，然后交了一个感觉没错的背包上去。 结果无情30分。 看了一下数据范围，再仔细想了下，发现不是dp，贪心就可以了，从小到大排序费用，再从小到大买，到买不起为止即可。 Code:
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摘要："机器分配" Code: c++ include include include using namespace std; //Mystery_Sky // define M 5000 int f[M][M]; int n, m, ans; int c[M][M]; void print(int i
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摘要："开餐馆" 01背包的变式 Code:
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摘要："分组背包" 分组背包模板题 Code:
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摘要："宠物小精灵之收服" 二维费用背包 Code: c++ include include include include using namespace std; //Mystery_Sky // define M 5000 int f[M][M]; int n, m, k; int a[M], b[
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摘要："潜水员" 二维费用背包 Code:
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摘要："Another One Bites The Dust" CF的题目在你谷上难度虚高似乎已成常态 不过这道题相比于~~愚人节的那几道~~相对好得多，没有被评成紫题。 这道题题面意思比较清楚，就是对于给定数量的'a'、'b'、'ab',求出能组成的最长‘ab'交替串。 设有x个'a',y个'b',
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