LuoguP1858 多人背包(DP)

Posted 2020-12-02 bingoyes

第(K)优解这类问题可在(DP)过程中通过添维解决。归并出当前前(K)大的解。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++a)
#define nR(a,b,c) for(register int a = (b); a >= (c); --a)
#define Fill(a,b) memset(a, b, sizeof(a))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
#define QWQ
#ifdef QWQ
#define D_e_Line printf("
---------------
")
#define D_e(x) cout << (#x) << " : " << x << "
"
#define Pause() system("pause")
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out.txt", "w", stdout)
#define TIME() fprintf(stderr, "
TIME : %.3lfms
", clock() * 1000.0 / CLOCKS_PER_SEC)
#else
#define D_e_Line ;
#define D_e(x) ;
#define Pause() ;
#define FileOpen() ;
#define FileSave() ;
#define TIME() ;
#endif
struct ios {
    template<typename ATP> inline ios& operator >> (ATP &x) {
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
        while(c >= '0' && c <='9') x = x * 10 + (c ^ '0'), c = getchar();
        x *= f;
        return *this;
    }
}io;
using namespace std;
template<typename ATP> inline ATP Max(ATP a, ATP b) {
    return a > b ? a : b;
}
template<typename ATP> inline ATP Min(ATP a, ATP b) {
    return a < b ? a : b;
}
template<typename ATP> inline ATP Abs(ATP a) {
    return a < 0 ? -a : a;
}

const int N = 207;

long long f[5007][51], tmp[5007];

struct nod {
    long long cost, val;
} a[N];
int main() {
    int n, K, V;
    io >> K >> V >> n;
    R(i,1,n){
        io >> a[i].cost >> a[i].val;
    }
    
    
    Fill(f, 0xcf);
    f[0][1] = 0;
    
    R(i,1,n){
        nR(j,V,a[i].cost){
            int tot1 = 1, tot2 = 1, tot3 = 0;
            while(tot3 <= K){
                if(f[j][tot1] > f[j - a[i].cost][tot2] + a[i].val){
                    tmp[++tot3] = f[j][tot1++];
                }
                else{
                    tmp[++tot3] = f[j - a[i].cost][tot2++] + a[i].val;
                }
            }
            R(k,1,K) f[j][k] = tmp[k];
        }
    }
    
    long long ans = 0;
    R(i,1,K){
        ans += f[V][i];
    }

    printf("%lld", ans);
    return 0;
}