UVA607 Scheduling Lectures

换行出错。WA了一次。。

用a[i][j]表示从第i个主题上到第j个主题所需要的时间，d[i][0]表示上完第i个主题所需课时数的最小值，d[i][1]表示在上完第i个主题所需课时数最小的情况下，不满意度DI的最小值。

下面贴代码：

#include <stdio.h>
#include <string.h>
int a[1001][1001],d[1001][2],C,time;
int DI(int a)
{
    if(a == time)
        return 0;
    if(time - a <= 10)
        return -C;
    return (time - a - 10) * (time - a - 10);
}
int main()
{
    int ntopics;
    int ncase = 0;
    while(scanf("%d",&ntopics) == 1)
    {
        if(ntopics == 0)
            break;
        scanf("%d",&time);
        scanf("%d",&C);
        int i,j,k,temp;
        //处理a[][]数组
        a[0][0] = 0;
        for(i = 1;i <= ntopics;i++)
        {
            scanf("%d",&temp);
            k = i;
            a[i - 1][i] = temp;
            for(j = i - 2;j >= 0;j--)
            {
                a[j][k] = temp + a[j][k - 1];
            }
        }
        /*for(i = 1;i <= ntopics;i++)
        {
            for(j = 0;j < i;j++)
                printf("%d  ",a[j][i]);
            printf("\n");
        }*/
        
        //DP
        d[0][0] = 0;
        d[0][1] = 0;
        for(i = 1;i <= ntopics;i++)
        {
            int temp1,temp2;
            temp1 = temp2 = 0xffffff;
            for(k = 0;k <= i - 1;k++)
            {
                if(a[k][i] <= time && d[k][0] + 1 <= temp1)
                {
                    temp1 = d[k][0] + 1;
                    if(d[k][1] + DI(a[k][i]) < temp2)
                        temp2 = d[k][1] + DI(a[k][i]);
                }
            }
            d[i][0] = temp1;
            d[i][1] = temp2;
        }
        if(ncase)
            printf("\n");
        printf("Case %d:\n",++ncase);
        printf("Minimum number of lectures: %d\n",d[ntopics][0]);
        printf("Total dissatisfaction index: %d\n",d[ntopics][1]);
    }
    return 0;
}