【bailian4102 】宠物小精灵之收服

最新推荐文章于 2023-02-22 21:44:56 发布

原创最新推荐文章于 2023-02-22 21:44:56 发布 · 194 阅读

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本文深入探讨了二维背包问题的解决方案，通过动态规划方法求解在有限容量下如何选择物品以达到最大价值。代码示例清晰展示了状态转移方程及边界条件设定。

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bailian4102
这是二维背包裸题吧
最后统计了一下最小的代价获得ans 那么M-代价就是我们花费的体积

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
const int MAX_N = 105;
int v[MAX_N],w[MAX_N],dp[1025][525];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int N,M,n;
    scanf("%d%d%d",&N,&M,&n);
    for(int i = 1;i<=n;++i) scanf("%d%d",&v[i],&w[i]);
    for(int i = 1;i<=n;++i)
    {
        for(int j = N;j>=v[i];j--)
        {
            for(int k = M;k>=w[i];k--)
            {
                dp[j][k] = max(dp[j][k],dp[j-v[i]][k-w[i]]+1);
            }
        }
    }
    int maxx = dp[N][M],ans= inf;
    for(int i = 0;i<=N;++i)
    {
        for(int j = 0;j<=M;++j)
        {
            if(dp[i][j]==maxx)
            {
                ans = min(ans,j);
            }
        }
    }
    printf("%d %d\n",maxx,M-ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}