hdu 2844 Coins(DP, 背包)

题意：硬币的面值 A1....An, 每种面值持有数 C1....Cn, 求 [1, m] 中能够凑出的面值数。

多重背包附加二进制优化buff。。。

Ai * Ci >= m 的情况用完全背包处理。。。

挺好的一道题，两种背包都考了还加个优化。。。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<map>
#include<algorithm>
#include<vector>
#include<queue>
#include<cstring>
#include<set>
#include<list>
#include<cmath>
using namespace std;

typedef long long LL;

const int maxn = 100;
const int inf = 0x7fffffff;

int a[maxn+5];
int c[maxn+5];
int f[100000+5];
vector<int> pack[maxn+5];
int n, m;

void do_pack(int idx) {
pack[idx].clear();
int num = c[idx];
if (num * a[idx] >= m) return;
int base = 1;
vector<int> tmp;
while (num >= base) {
tmp.push_back(base);
num -= base;
base <<= 1;
}
if (num) tmp.push_back(num);
int sz = tmp.size();
for (int i=sz-1;i>=0;--i)
pack[idx].push_back(tmp[i]);
}

void print(int * A, int n) {
for (int i=0;i<=n;++i) cout << A[i] << " ";cout << endl;
}

int main() {
//freopen("input.in", "r", stdin);
while (scanf("%d%d",&n,&m)) {
if (n == 0 && m == 0) break;
for (int i=1;i<=n;++i) scanf("%d",&a[i]);
for (int i=1;i<=n;++i) scanf("%d",&c[i]);
for (int i=1;i<=n;++i) do_pack(i);
//print(a, n);
//print(c, n);
#if 0
for (int i=1;i<=n;++i) {
cout << "pack " << i << ": ";
int sz = pack[i].size();
for (int j=0;j<sz;++j) cout << pack[i][j] << " ";cout << endl;
}
#endif
for (int i=0;i<=m;++i) f[i] = 0;
f[0] = 1;
for (int i=1;i<=n;++i) {
//cout << "fuck " << i << " size: " << pack[i][0] << endl;
if (c[i]*a[i] >= m) {
for (int j=a[i];j<=m;++j)
f[j] |= f[j-a[i]];
continue;
}
int sz = pack[i].size();
for (int k=0;k<sz;++k)
for (int j=m;j>=0;--j) {
if (f[j]) continue;
if (j-a[i]*pack[i][k] >= 0) {
//cout << "fuck k " << k << endl;
f[j] |= f[j-a[i]*pack[i][k]];
}
#if 0
print(f, m);
#endif
}

}
int ans = 0;
for (int i=1;i<=m;++i)
if (f[i]) ++ans;
printf("%d\n", ans);
}
return 0;
}

------------------------------------------------------------------------------

又写了一遍，发现 Ai*Ci>=m 时，不搞完全背包也能过。。。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <cassert>
#include <algorithm>
#include <cmath>
#include <limits>
#include <set>
#include <map>
using namespace std;

#define SPEED_UP iostream::sync_with_stdio(false);
#define FIXED_FLOAT cout.setf(ios::fixed, ios::floatfield);
#define rep(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
#define urep(i, s, t) for(int (i)=(s);(i)>=(t);--(i))

typedef long long LL;

const int Maxn = 100000;
struct Node{
int v, t;
};
int n, m, v[100+5], t[100+5], f[Maxn+5];
vector<Node> vec;

bool init() {
cin >> n >> m;
if (!n && !m) return false;
rep(i, 0, n-1) cin >> v[i];
rep(i, 0, n-1) cin >> t[i];

vec.clear();
rep(i, 0, n-1) {
int sz = 1, left = t[i];
while (left >= sz) {
vec.push_back((Node){v[i], sz});
left -= sz;
sz <<= 1;
}
if (left) vec.push_back((Node){v[i], left});
}

return true;
}

int main() {
#ifndef ONLINE_JUDGE
freopen("input.in", "r", stdin);
#endif
SPEED_UP

while (init()) {
int sz = vec.size();
memset(f, 0, sizeof(f));
f[0] = 1;
rep(i, 0, sz-1) {
int _v = vec[i].v * vec[i].t;
urep(j, m, _v) f[j] |= f[j-_v];
}
int cnt = 0;
rep(i, 1, m) cnt += f[i];
cout << cnt << endl;
}

return 0;
}