xutian_curry的博客

简化版的01背包#include <iostream>#include <stdio.h>#include <algorithm>#include <vector>#include <string>#include <map>#include <set>#include <queue&a

完全背包中每种物品的数量是无穷多个。与01背包差别的地方，更新dp[i][j]是顺序更新，因为每个背包可以无限次选择，状态 dp[ i ][ j ]恰好可以由可能已经放入物品i的状态 dp[ i ] [ j - list[i].w]转换过来#include <iostream>#include <memory.h>using namespace std;int n,m...

f(i,j)=前i个糖果选一些构成j%k的最大值函数里的j是mod k过的，方程为f(i,j)=max(f(i,j),f(i-1,(j-a[i]%k+k)%k)+a[i]; 边界为d[i][0]=0。答案即为dp[n][0]。#include <iostream>#include <memory.h>#include <string>#include &l...

裸的二维费用背包#include <iostream>#include<cstring>#include<algorithm>using namespace std;int n,m,k;int a[1005];//消耗精灵球int b[1005];int result[1005][1005];int main(){ cin >&g...

#include <iostream>#include <stdio.h>#include <algorithm>#include <vector>#include <string>#include <map>#include <set>#include <queue>#

背包问题#include <iostream>#include <queue>#include <memory.h>#include <stdio.h>#include <stdlib.h>#include <math.h>#include <string>#define eps 1e-6#defi