Shooting Contest
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 4127 | Accepted: 1516 | Special Judge |
Description
Welcome to the Annual Byteland Shooting Contest. Each competitor will shoot to a target which is a rectangular grid. The target consists of r*c squares located in r rows and c columns. The squares are coloured white or black. There are exactly two white squares and r-2 black squares in each column. Rows are consecutively labelled 1,..,r from top to bottom and columns are labelled 1,..,c from left to right. The shooter has c shots.
A volley of c shots is correct if exactly one white square is hit in each column and there is no row without white square being hit. Help the shooter to find a correct volley of hits if such a volley exists.
Example
Consider the following target:
Volley of hits at white squares in rows 2, 3, 1, 4 in consecutive columns 1, 2, 3, 4 is correct.
Write a program that: verifies whether any correct volley of hits exists and if so, finds one of them.
Input
The first line of the input contains the number of data blocks x, 1 <= x <= 5. The following lines constitute x blocks. The first block starts in the second line of the input file; each next block starts directly after the previous one.
The first line of each block contains two integers r and c separated by a single space, 2 <= r <= c <= 1000. These are the numbers of rows and columns, respectively. Each of the next c lines in the block contains two integers separated by a single space. The integers in the input line i + 1 in the block, 1 <= i <= c, are labels of rows with white squares in the i-th column.
Output
For the i-th block, 1 <= i <= x, your program should write to the i-th line of the standard output either a sequence of c row labels (separated by single spaces) forming a correct volley of hits at white squares in consecutive columns 1, 2, ..., c, or one word NO if such a volley does not exists.
Sample Input
2
4 4
2 4
3 4
1 3
1 4
5 5
1 5
2 4
3 4
2 4
2 3
Sample Output
2 3 1 4
NO
題目大意:給你一個r*c的矩陣,然後這個矩陣每一列有兩個格子是白色的,然後你在每列上開一槍,可以打這列的任意的白色格子,問你能否消除全部的行,如果包括所有行輸出你在1-c列打搶的格子的行,當然這可以是不唯一的,記好!!!是不唯一!!!!
如果打不到就輸出NO
題目首先給出有幾組資料,對于每組資料第一行代表r和c,接下來c行有兩個數,代表這行的哪兩列是白色格子
解題思路:和以前那個經典的二分比對差不多,就是在一行打一槍可以消滅所有行那個題目,很容易想到二分比對,我們将白色格子的行指向列,然後求最大比對,用行去比對列,看是否能夠得到的比對數是r,如果是r的話則代表所有的行都可以被打到,然後每列對應的行,如果比對數不是r就輸出NO
當然有特例,如果r>c,這種情況開c搶根本不可能達到r行,是以直接輸出NO
代碼:
#include<stdio.h>
#include<string.h>
bool map[1005][1005];
bool used[1005];
int link[1005];
int n,m;
bool dfs(int u)//尋找增廣路! 找到傳回1,否則傳回0!
{
int i;
for(i=1;i<=m;i++)
{
if(map[u][i]&&!used[i])])//與x相連點,并且沒有周遊到
{
used[i]=true;;//标記為周遊過
if(link[i]==-1||dfs(link[i]))//i 點 沒有和另一部分比對或 和i配對的點 沒有比對!
{
link[i]=u;
return true;
}
}
}
return false;
}
int main()
{
int i,j,x,y,t,num;
scanf("%d",&t);
while(t--)
{
num=0;
scanf("%d%d",&n,&m);
memset(map,0,sizeof(map));
memset(link,-1,sizeof(link));
if(n>m)//行大于列
{
printf("NO\n");
continue;
}
for(i=1;i<=m;i++)//建圖
{
scanf("%d%d",&x,&y);
map[x][i]=true;
map[y][i]=true;
}
for(i=1;i<=n;i++)
{
memset(used,0,sizeof(used));
if(dfs(i))
num++;
}
if(num<n)//不是完美比對
{
printf("NO\n");
continue;
}
else
{
for(i=1;i<=m;i++)
{
if(link[i]!=-1)
printf("%d ",link[i]);
else
{
for(j=1;j<=n;j++)
if(map[j][i])
{
printf("%d ",j);
break;
}
}
}
printf("\n");
}
}
return 0;
}