UVA 624 CD

UVA_624

    实际就是0-1背包的问题，把时间等效成体积和价值就可以了。需要注意的是总时间的那个维度要开大一点，10000就够用了。

#include<stdio.h>
#include<string.h>
#define MAXN 30
#define MAXD 10010
int M, N, a[MAXN], p[MAXN][MAXD], f[MAXN][MAXD];
void init()
{
    int i;
    scanf("%d", &N);
    for(i = 1; i <= N; i ++)
        scanf("%d", &a[i]);
}
void printresult(int i, int j)
{
    if(i == 0)
        return;
    printresult(i - 1, p[i][j]);
    if(p[i][j] < j)
        printf("%d ", a[i]);
}
void solve()
{
    int i, j, temp;
    memset(f, 0, sizeof(f));
    for(i = 1; i <= N; i ++)
        for(j = 0; j <= M; j ++)
        {
            f[i][j] = f[i - 1][j];
            p[i][j] = j;
            if(j >= a[i])
            {
                temp = f[i - 1][j - a[i]] + a[i];
                if(temp > f[i][j])
                {
                    p[i][j] = j - a[i];
                    f[i][j] = temp;
                }
            }
        }
    printresult(N, M);
    printf("sum:%d\n", f[N][M]);
}
int main()
{
    while(scanf("%d", &M) == 1)
    {
        init();
        solve();
    }
    return 0;
}