嘟嘟嘟
既然让求前(k)优解,那么就多加一维,(dp[j][k])表示体积为(j)的第(k)优解是啥((i)一维已经优化掉了)。
考虑原来的转移方程:dp[j] = max(dp[j], dp[j - c[i]] + v[i])。
现在多了一维,那么dp‘[j][k]就分别从dp[j][]和dp[j - c[i]][]中取前(k)大的即可。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxm = 5e3 + 5;
const int maxk = 55;
const int maxn = 205;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
int n, m, K, c[maxn], v[maxn];
int dp[maxm][maxk];
int main()
{
K = read(); m = read(); n = read();
for(int i = 1; i <= n; ++i) c[i] = read(), v[i] = read();
Mem(dp, 128); dp[0][1] = 0;
for(int i = 1; i <= n; ++i)
for(int j = m; j >= c[i]; --j)
{
int t1 = 1, t2 = 1, len = 0;
static int tp[maxk];
while(len < K)
{
if(dp[j][t1] > dp[j - c[i]][t2] + v[i]) tp[++len] = dp[j][t1++];
else tp[++len] = dp[j - c[i]][t2++] + v[i];
}
for(int k = 1; k <= K; ++k) dp[j][k] = tp[k];
}
int ans = 0;
for(int i = 1; i <= K; ++i) ans += dp[m][i];
write(ans), enter;
}