poj 1719 Shooting Contest

http://poj.org/problem?id=1719

Shooting Contest

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4135		Accepted: 1521		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

题意比较难理解，题意弄懂后这道题就比较简单，套用匈牙利算法求最大匹配

题目大意：r*c的矩阵，矩阵由白格子和黑格子组成，每一列有两个格子是白色的剩下的为黑色，每一列射击一发子弹击中白色格子，问是否所有行都有白色格子被击中

数据分析：

2//数据组数
4 4//行r  列c
2 4//第1列的第2行和第4行是白色格子
3 4//第2列的第3行和第4行是白色格子
1 3//第3列的第1行和第3行是白色格子
1 4// ...

1.如果 r > c , c列射击完后仍会有行没有被射击过；
2.要从每一列开始射击并射中白色格子，即将行r和列c作为X，Y集合，白色格子部分进行匹配，得到最大匹配值ans
    1>如果ans==r
                 （1）如果每一列都有匹配（即都能射中某一行的白色格子）就输出与该列匹配的行（即该列击中的的白色格子所在的行）
                 (2)如果某一列没有找到匹配的格子，那个只要在该行任意选择一个白色的格子就可以了，输出该白色格子所在的行  
    2>如果ans>r或者ans<r都无法满足每行都被射击过

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<queue>
#include<algorithm>
#define INF 0x3f3f3f3f
#define N 1010

using namespace std;

int G[N][N], vis[N], used[N];
int r, c;

bool Find(int u)
{
    int i;
    for(i =  ; i <= c ; i++)
    {
        if(!vis[i] && G[u][i])
        {
            vis[i] = ;
            if(!used[i] || Find(used[i]))
            {
                used[i] = u;
                return true;
            }
        }
    }
    return false;
}//匈牙利

int main()
{
    int a, b, i, t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &r, &c);
        if(r > c)
        {
            printf("NO\n");
            continue;
        }//如果 r > c , c列射击完后仍会有行没有被射击过
        memset(G, , sizeof(G));
        for(i =  ; i <= c ; i++)
        {
            scanf("%d%d", &a, &b);
            G[a][i] = G[b][i] = ;//第i列个第a行和第b行是白色格子
        }
        memset(used, , sizeof(used));
        int ans = ;
        for(i =  ; i <= r ; i++)
        {
            memset(vis, , sizeof(vis));
            if(Find(i))
                ans++;
        }
        if(ans == r)
        {
            for(i =  ; i <= c ; i++)
            {
                if(used[i] != )//如果每一列都有匹配（即都能射中某一行的白色格子）就输出与该列匹配的行（即该列击中的的白色格子所在的行）
                    printf("%d ", used[i]);
                else
                {
                    for(int j =  ; j <= r ; j++)
                    {
                        if(G[j][i])
                        {
                            printf("%d ", j);
                            break;
                        }
                    }
                }//如果某一列没有找到匹配的格子，那个只要在该行任意选择一个白色的格子就可以了，输出该白色格子所在的行；
            }
             printf("\n");
        }
        else//如果ans>r或者ans<r都无法满足每行都被射击过
            printf("NO\n");
    }
    return ;
}