poj 2584 T-Shirt Gumbo (二分匹配)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 2571 | Accepted: 1202 |
Description
Input
A single data set has 4 components:
- Start line - A single line:
START Xwhere (1 <= X <= 20) is the number of contestants demanding shirts.
- Tolerance line - A single line containing X space-separated pairs of letters indicating the size tolerances of each contestant. Valid size letters are S - small, M - medium, L - large, X - extra large, T - extra extra large. Each letter pair will indicate the range of sizes that will satisfy a particular contestant. The pair will begin with the smallest size the contestant will accept and end with the largest. For example:
MXwould indicate a contestant that would accept a medium, large, or extra large T-shirt. If a contestant is very picky, both letters in the pair may be the same.
- Inventory line - A single line:
S M L X Tindicating the number of each size shirt in Boudreaux and Thibodeaux's inventory. These values will be between 0 and 20 inclusive.
- End line - A single line:
END
After the last data set, there will be a single line:
ENDOFINPUT
Output
T-shirts rock!
Otherwise, output:
I'd rather not wear a shirt anyway...
Sample Input
START 1
ST
0 0 1 0 0
END
START 2
SS TT
0 0 1 0 0
END
START 4
SM ML LX XT
0 1 1 1 0
END
ENDOFINPUT
Sample Output
T-shirts rock!
I'd rather not wear a shirt anyway...
I'd rather not wear a shirt anyway...
Source
//140K 0MS C++ 1519B 2014-06-09 08:53:58
/*
题意:
有x个人,没个人要穿的的衣服码数要在一个范围内,给出5种不同码的衣服的数量,问是否可以满足x个人的需求。 最大匹配:
构好图后直接最大匹配即可。构图的要知道是什么和什么匹配,要以人和衣服匹配,人数是固定的,衣服也是,
即是二分图的两个集合,匹配时每一件衣服作为一个点,而不是每一类衣服作为一个点。 */
#include<stdio.h>
#include<string.h>
int g[][];
int match[];
int vis[];
char r[]={"SMLXT"};
int x;
int judge(int i,char range[])
{
int lr,rr;
for(int j=;j<;j++){
if(range[]==r[j]) lr=j;
if(range[]==r[j]) rr=j;
}
if(i>=lr && i<=rr) return ;
return ;
}
int dfs(int u)
{
for(int i=;i<x;i++)
if(!vis[i] && g[u][i]){
vis[i]=;
if(match[i]==- || dfs(match[i])){
match[i]=u;
return ;
}
}
return ;
}
int hungary(int pos)
{
int ret=;
memset(match,-,sizeof(match));
for(int i=;i<=pos;i++){
memset(vis,,sizeof(vis));
ret+=dfs(i);
}
return ret;
}
int main(void)
{
char op[],stu[][];
int a[];
while(scanf("%s",op)!=EOF)
{
if(strcmp(op,"ENDOFINPUT")==) break;
scanf("%d",&x);
for(int i=;i<x;i++)
scanf("%s",&stu[i]);
for(int i=;i<;i++)
scanf("%d",&a[i]);
scanf("%s",op);
memset(g,,sizeof(g));
int pos=;
for(int i=;i<;i++){
while(a[i]--){
pos++;
for(int j=;j<x;j++)
if(judge(i,stu[j]))
g[pos][j]=;
}
}
if(hungary(pos)==x)
puts("T-shirts rock!");
else puts("I'd rather not wear a shirt anyway...");
}
return ;
}
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