思路:明显的背包问题,并且是多重背包,多重背包有两种思路
思路1:当成一种新情况考虑,因为有个数的限制所以跟完全背包不一样的地方在于状态转移方程应该是dp[i]=max{dp[i],dp[i-k*w[i]+k*v[i]}(k:0->j/w[i]),但是这个会超时,因为时间复杂度是n*w*j/[w[i]];
附上一段超时的代码:
#include"stdio.h"
#include <iostream>
#include<stdlib.h>
#include<algorithm>
#include<set>
#include<time.h>
#include<string>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
using namespace std;
#define maxn 500005
long dp[maxn];
long v[maxn], w[maxn],c[maxn];
int main()
{
int n, W;
while (cin>>n>>W)
{
memset(dp, 0, sizeof(dp));
memset(v, 0, sizeof(v));
memset(w, 0, sizeof(w));
memset(c, 0, sizeof(c));
for (int i = 1; i <= n; i++)
{
cin >> w[i] >> v[i] >> c[i];
for (int j = W; j >= w[i]; j--)
{
long maxm = dp[j];
for (int k = 0; k <= j / w[i] && k <= c[i]; k++)
{
if (dp[j - k * w[i]] + v[i] * k > maxm)
{
maxm = dp[j - k * w[i]] + v[i] * k;
dp[j] = maxm;
}
}
}
}
cout << dp[W] << endl;
}
return 0;
}
思路2:把多重背包化简成01背包,但是这种思路有一个问题,就是会超时,所以要用2进制进行优化(如有同种物品有7个,则用1,2,4进行组合就能得到1-7中所有的情况,如果同种物品有10个,则用1,2,4组合加上剩下的3进行组合就能得到所有情况)
01背包的状态转移方程是dp[i]=max{dp[i],dp[i-w[i]]+v[i]},这个方法的时间复杂度是n*log(count);
附上ac代码:
#include"stdio.h"
#include <iostream>
#include<stdlib.h>
#include<algorithm>
#include<set>
#include<time.h>
#include<string>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
using namespace std;
#define maxn 500005
long dp[maxn];
long v[maxn], w[maxn];
int main()
{
int n, W;
while (cin>>n>>W)
{
memset(dp, 0, sizeof(dp));
memset(v, 0, sizeof(v));
memset(w, 0, sizeof(w));
int a,b,count;
long kind = 1;
for (int i = 0; i < n; i++)
{
cin >> a >> b >> count;
int j;
for (j = 1; j <= count; j <<= 1)
{
v[kind] = j * b;
w[kind] = j * a;
count -= j;
kind++;
}
if (count)
{
v[kind] = (count)*b;
w[kind] = (count)*a;
kind++;
}
}
for (int i = 1; i <= kind; i++)
{
for (int j = W; j >= w[i]; j--)
{
dp[j] = max(dp[j], dp[j - w[i]] + v[i]);
}
}
cout << dp[W] << endl;
}
return 0;
}