<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">01揹包問題:</span>
01揹包問題是說有一個重量爲w的包,有n個物體,第i個物體的重量是w[I],價值是V[I],每個物體最多隻能拿一次,每個物體只有一個,怎樣儘可能地裝滿包,使得價值最大,並且總的體重不超過w;
分析:之所以稱爲01揹包是因爲物體可以拿也可以不拿。那麼分析一下狀態,我們需要知道取了第幾個物體,以及此時的揹包大小最多可裝多少價值的物體。
dp[i][j]:i表示取到第i個物體;j表示揹包容量爲j;dp[i][j]表示最大價值。
狀態轉移:取到第i個物體揹包容量可以裝下時,我們可以要,也可以不要,什麼時候要這個物體?
————>當要第i個物體時:dp[i][j]=dp[i-1][j-w[i]]+v[i];(爲什麼是j-w[i]?模擬一下揹包放物體的過程我們就知道,當要放第i個物體時,揹包容量就爲j-w[i],剩餘揹包儘可能多地裝前面i-1個物體。)
————>當第i個物體不裝時,那麼dp[i][j]=dp[i-1][j];
取兩者中的最大值。得到的便是最優解。
例題:
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 29060 Accepted Submission(s): 11871
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=1005;
int dp[maxn][maxn];
int va[maxn];
int w[maxn];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));
int n,v;
scanf("%d%d",&n,&v);
for(int i=1;i<=n;i++)
scanf("%d",&va[i]);
for(int i=1;i<=n;i++)
scanf("%d",&w[i]);
for(int i=1;i<=n;i++)
{
for(int j=0;j<=v;j++)
{
if(j<w[i])
dp[i][j]=dp[i-1][j];//如果揹包的容量比第i個物體小,放不下這個物體,相當於不放。
else
dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+va[i]);//狀態轉移。
}
}
printf("%d\n",dp[n][v]);
}
}
優化如下:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=1005;
int dp[maxn];//容量爲j的揹包能放的最大價值
int va[maxn];
int w[maxn];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));
int n,v;
scanf("%d%d",&n,&v);
for(int i=1;i<=n;i++)
scanf("%d",&va[i]);
for(int i=1;i<=n;i++)
scanf("%d",&w[i]);
for(int i=1;i<=n;i++)
{
for(int j=v;j>=w[i];j--)//j從最大更新,只需要更新到w[i],因爲當揹包裝不下i時無需更新,已經是最大了。
{
dp[j]=max(dp[j],dp[j-w[i]]+va[i]);
}
}
printf("%d\n",dp[v]);
}
}