大0-1背包

Beautiful Land

#include<cstdio>价值背包
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<map>
#define max(a,b)(a>b?a:b)
#define min(a,b)(a<b?a:b)
typedef long long ll;
using namespace std;
#define N 10005
ll dp[N],w[N];  ///此时的dp[i]表示的是；价值为i时的最小容量为dp[i];
int v[N];
  
int main()
{
    int T,i,n,sum;
    ll V;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%I64d",&n,&V);
        sum=0;
        for(i=1;i<=n;i++)
        {
            scanf("%I64d%d",&w[i],&v[i]);
            sum=sum+v[i];
        }
  
        memset(dp,1000000010,sizeof(dp)); ///要求最小容量，初始化为最大值；
        dp[0]=0;
        for(i=1;i<=n;i++)
        {
            for(int j=sum;j>=v[i];j--)
                dp[j]=min(dp[j],dp[j-v[i]]+w[i]);
        }
  
        for(i=sum;i>=0;i--)
        {
            if(dp[i]<=V)
            {
               printf("%d\n",i); ///此处输出i，即为满足条件的最大价值
               break;
            }
        }
    }
    return 0;
}

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卡迪亚的游戏

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <deque>
#include <queue>
#include <stack>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <iomanip>
#define ull unsigned long long
#define ll long long
#define pb push_back
#define mem(sum,x) memset(sum,x,sizeof(sum))
#define rep(i,start,end) for(int i=start;i<=end;i++)
#define per(i,end,start) for(int i=end;i>=start;i--)
#define tle ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
using namespace std;
const int mod = 998244353;
const int mxn = 2e4 +7;
int _,n,m,k,t,u,ans,cnt,ok,lim;
ll w[mxn] , cost[mxn] , dp[mxn];
#define lc now<<1
#define rc now<<1|1
bool vis[mxn];
int main()
{
    ll lim , cnt = 1 , ans = 0  ;
    scanf("%d %lld",&n,&lim);
    for(int i=1;i<=n;i++)
        scanf("%lld %lld",&cost[i],&w[i]) , ans += w[i];
    for(int i=0;i<=ans;i++)
        dp[i] = mod ;
    dp[0] = 0 ;
    for(int i=1;i<=n;i++)
    {
        for(int j=ans;j>=w[i];j--)
        {
            dp[j] = min(dp[j] , dp[j- w[i] ]+cost[i]);
        }
    }
    for(int i=ans;i>=0;i--)
    {
        if(dp[i]<=lim)
        {
            printf("%d\n",i);
            break;
        }
    }

}

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Knapsack problem