完全背包题目：

pku 1384 Piggy-Bank 完全背包入门题目。 

http://poj.org/problem?id=1384

这里只是求的恰好装满，且是最小罢了。在恰好装满时只要给f[0] = 0; 其他的一个未定义状态负无穷正无穷即可。

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a , b) ((a) < (b) ? (a) : (b))
#define Max(a , b) ((a) > (b) ? (a) : (b))

#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 10007
#define N 504
using namespace std;
//freopen("din.txt","r",stdin);

int f[M],w[N],c[N];

int main(){
    //freopen("din.txt","r",stdin);
    int n,m;
    int T,i,j;
    scanf("%d",&T);
    while (T--){
        scanf("%d%d",&n,&m);
        int V = m - n;
        scanf("%d",&n);
        for (i = 0; i < n; ++i){
            scanf("%d%d",&w[i],&c[i]);
        }

        for (i = 0; i < M; ++i) f[i] = inf;
        f[0] = 0;

        for (i = 0; i < n; ++i){
            for (j = c[i]; j <= 10000; ++j){
                f[j] = min(f[j],f[j - c[i]] + w[i]);
            }
        }
        if (f[V] != inf) printf("The minimum amount of money in the piggy-bank is %d.\n",f[V]);
        else printf("This is impossible.\n");
    }
    return 0;
}

 

pku 2063 Investment 多次完全背包

http://poj.org/problem?id=2063

题意：

给定初始金钱数，给出d种债券的利润 c[i],w[i]每年买c[i]的债券可得w[i]的利润。给定你的使用年份m，问我利用初始给定的钱数最后得到的最大总钱数。

思路；

将利润看做物品的w值，将债券钱数看做物品的花费。往背包容量为初始钱数的背包里面放物品所得的最大值（最大利润）这样求即可。。

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a , b) ((a) < (b) ? (a) : (b))
#define Max(a , b) ((a) > (b) ? (a) : (b))

#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 100007
#define N 17
using namespace std;
//freopen("din.txt","r",stdin);

int f[M];
int c[N],w[N];

int main(){
    //freopen("din.txt","r",stdin);
    int i,j,T,d;
    int n,m;
    scanf("%d",&T);
    while (T--){
        scanf("%d%d",&n,&m);
        scanf("%d",&d);
        for (i = 1; i <= d; ++i){
            scanf("%d%d",&c[i],&w[i]);
            c[i] /= 1000;
        }
        while (m--){
            for (i = 0; i < M; ++i) f[i] = 0;
            int tmp = n/1000;
            for (i = 1; i <= d; ++i){
                for (j = c[i]; j <= tmp; ++j){
                    f[j] = max(f[j],f[j - c[i]] + w[i]);
                }
            }
            n += f[tmp];
        }
        printf("%d\n",n);
    }
    return 0;
}

 

 

待更新。。。

转载于:https://www.cnblogs.com/E-star/archive/2012/10/10/2717831.html