> 算法思想是这样的,对S个人的要求按最小的size进行排序,然后顺序扫描求解
> #include<ctype.h>
> #include<stdio.h>
> struct node
> {
> int x;
> int y;
> };
>  
> main()
> {
> char ch[30],c[21][2];
> int m,min,t,flag,sum,a[6],s,i,j;
> struct node no[30];
> long k;
> for (k=0;;k++)
> {
> scanf("%s",ch);
> if (!strcmp(ch,"ENDOFINPUT"))
> break;
> sum=0;
> scanf("%d",&s);
> for (i=1;i<=s;i++)
> scanf("%s",c[i]);
>  
> for (i=1;i<=5;i++)
> {
> scanf("%d",&a[i]);
> sum+=a[i];
> }
>  
> if (sum<s)
> {
> scanf("%s",ch);
> printf("I'd rather not wear a shirt anyway...\n");
> continue;
> }
>  
> for (i=1;i<=s;i++)
> {
> switch (c[i][0])
> {
> case 'S':no[i].x=1;break;
> case 'M':no[i].x=2;break;
> case 'L':no[i].x=3;break;
> case 'X':no[i].x=4;break;
> case 'T':no[i].x=5;break;
> }
> switch (c[i][1])
> {
> case 'S':no[i].y=1;break;
> case 'M':no[i].y=2;break;
> case 'L':no[i].y=3;break;
> case 'X':no[i].y=4;break;
> case 'T':no[i].y=5;break;
> }
> }
>  
> for (i=1;i<s;i++)   //对每个人的要求按最小的号进行排序
> {
> m=i;
> min=no[i].x;
> for (j=i+1;j<=s;j++)
> if (no[j].x<min)
> {
> m=j;
> min=no[j].x;
> }
> if (m!=i)
> {
> t=no[i].x;
> no[i].x=no[m].x;
> no[m].x=t;
> t=no[i].y;
> no[i].y=no[m].y;
> no[m].y=t;
> }
> }
>  
> flag=0;               //判断是否满足条件
> for (i=1;i<=s;i++)
> for (j=no[i].x;j<=no[i].y;j++)
> if (a[j])
> {
> a[j]--;
> flag++;
> break;
> }
>  
> if (flag!=s)
> {
> scanf("%s",ch);
> printf("I'd rather not wear a shirt anyway...\n");
> }
> else
> {
> scanf("%s",ch);
> printf("T-shirts rock!\n");
> }
> }
> }

• 能给出几组特殊的测试数据吗? (0) huicpc21 2005-08-18 11:58:56
• Re:用网络流吧 (36) jinmingyan 2005-08-18 16:58:02
  • 谢谢了,我再去试试看 (40) huicpc21 2005-08-18 17:04:00