网络流 -- Shooting Contest(二分匹配)

Shooting Contest

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2279		Accepted: 809		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

Source

CEOI 1997

/*
一开始拿到这道题一下子就想到了二分图的最大匹配，然后想来想去没有把图的映射关系
想明白，就放弃二分了，转去想DP，但是想来想去也没有找到子结构关系，于是又转回二
分图。经过仔细分析终于找到了规律，首先如果列数少于行数是肯定要输出NO的，因为一
列只能覆盖一行，所以不能覆盖所有行。接下来就是建图了，以行为左，列为右建立二分
图，然后对二分图算最大匹配total，如果total < 行数则输出NO，否则依次遍历所有列并
输出其对应的行即pre，对于pre[i] != 0的就输出pre[i]否则，从i可以连的行中随便找一个输出
即可
*/
#include <iostream>   
#define MAX_N 1000   
using namespace std;
bool v[MAX_N + 1];
int pre[MAX_N + 1];
int rowCol[MAX_N + 1][MAX_N + 1];
int colRow[MAX_N + 1];
int r, c;
bool canGo(int from)
{
<span style="white-space:pre">	</span>int num = rowCol[from][0];
<span style="white-space:pre">	</span>int p, to;
<span style="white-space:pre">	</span>for (p = 1; p <= num; p++)
<span style="white-space:pre">	</span>{
<span style="white-space:pre">		</span>to = rowCol[from][p];
<span style="white-space:pre">		</span>if (!v[to])
<span style="white-space:pre">		</span>{
<span style="white-space:pre">			</span>v[to] = true;
<span style="white-space:pre">			</span>if (pre[to] == 0 || canGo(pre[to]))
<span style="white-space:pre">			</span>{
<span style="white-space:pre">				</span>pre[to] = from;
<span style="white-space:pre">				</span>return true;
<span style="white-space:pre">			</span>}
<span style="white-space:pre">		</span>}
<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>return false;
}
int main()
{
<span style="white-space:pre">	</span>int caseN, i;
<span style="white-space:pre">	</span>FILE *file = freopen("in13.txt", "r", stdin);
<span style="white-space:pre">	</span>if (!file)
<span style="white-space:pre">		</span>return 1;
<span style="white-space:pre">	</span>scanf("%d", &caseN);
<span style="white-space:pre">	</span>while (caseN--)
<span style="white-space:pre">	</span>{
<span style="white-space:pre">		</span>int row1, row2;
<span style="white-space:pre">		</span>memset(rowCol, 0, sizeof(rowCol));
<span style="white-space:pre">		</span>scanf("%d%d", &r, &c);
<span style="white-space:pre">		</span>for (i = 1; i <= c; i++)
<span style="white-space:pre">		</span>{
<span style="white-space:pre">			</span>scanf("%d%d", &row1, &row2);
<span style="white-space:pre">			</span>rowCol[row1][0]++;
<span style="white-space:pre">			</span>rowCol[row1][rowCol[row1][0]] = i;
<span style="white-space:pre">			</span>colRow[i] = row1;
<span style="white-space:pre">			</span>rowCol[row2][0]++;
<span style="white-space:pre">			</span>rowCol[row2][rowCol[row2][0]] = i;
<span style="white-space:pre">		</span>}
<span style="white-space:pre">		</span>int total = 0;
<span style="white-space:pre">		</span>bool can = true;
<span style="white-space:pre">		</span>if (c < r) can = false;
<span style="white-space:pre">		</span>else
<span style="white-space:pre">		</span>{
<span style="white-space:pre">			</span>memset(pre, 0, sizeof(pre));
<span style="white-space:pre">			</span>for (i = 1; i <= r; i++)
<span style="white-space:pre">			</span>{
<span style="white-space:pre">				</span>memset(v, 0, sizeof(v));
<span style="white-space:pre">				</span>if (canGo(i)) total++;
<span style="white-space:pre">			</span>}
<span style="white-space:pre">			</span>if (total < r) can = false;
<span style="white-space:pre">		</span>}
<span style="white-space:pre">		</span>if (!can) printf("NO\n");
<span style="white-space:pre">		</span>else
<span style="white-space:pre">		</span>{
<span style="white-space:pre">			</span>for (i = 1; i <= c; i++)
<span style="white-space:pre">			</span>{
<span style="white-space:pre">				</span>if (i != 1) printf(" ");
<span style="white-space:pre">				</span>if (pre[i] != 0) printf("%d", pre[i]);
<span style="white-space:pre">				</span>else printf("%d", colRow[i]);
<span style="white-space:pre">			</span>}
<span style="white-space:pre">			</span>printf("\n");
<span style="white-space:pre">		</span>}


<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>return 0;
}