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");
}
}
}

• Re:郁闷,一直都是WA,请牛人帮忙看下呀(code)真不知道怎么办了 (6) huicpc21 2005-08-18 11:14:22
• Re:用网络流吧 (1588) lookus 2005-08-18 11:37:01
  • 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