HDU2819（二分图匹配，记录过程）

Swap

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2236    Accepted Submission(s): 801
Special Judge

Problem Description

Given an N*N matrix with each entry equal to 0 or 1. You can swap any two rows or any two columns. Can you find a way to make all the diagonal entries equal to 1?

 

Input

There are several test cases in the input. The first line of each test case is an integer N (1 <= N <= 100). Then N lines follow, each contains N numbers (0 or 1), separating by space, indicating the N*N matrix.

 

Output

For each test case, the first line contain the number of swaps M. Then M lines follow, whose format is “R a b” or “C a b”, indicating swapping the row a and row b, or swapping the column a and column b. (1 <= a, b <= N). Any correct answer will be accepted, but M should be more than 1000.

If it is impossible to make all the diagonal entries equal to 1, output only one one containing “-1”. 

 

Sample Input

2

0 1

1 0

2

1 0

1 0

 

Sample Output

1

R 1 2

-1

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

很好的一道题，值得深思。

题意：可交换任意两行或任意两列，最终是主对角线上全为1，输出交换过程

如果可行的话，一定可以只交换列或只交换行得到，因为不管怎么交换，原先在同一行的始终在同一行，原先在同一列的始终在同一列。想到这，构图也就出来了，对行和列构图，如果mp[i][j]==1则在i和j之间加边

/*
ID: LinKArftc
PROG: 2819.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

const int maxn = 110;

int mp[maxn][maxn];
int linker[maxn];
bool vis[maxn];
int n;

bool dfs(int u) {
    for (int v = 1; v <= n; v ++) {
        if (!vis[v] && mp[u][v]) {
            vis[v] = true;
            if (linker[v] == -1 || dfs(linker[v])) {
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}

int hungry() {
    memset(linker, -1, sizeof(linker));
    int ret = 0;
    for (int i = 1; i <= n; i ++) {
        memset(vis, 0, sizeof(vis));
        if (dfs(i)) ret ++;
    }
    return ret;
}

int a[maxn], b[maxn];

int main() {
    //input;
    while (~scanf("%d", &n)) {
        int tmp;
        memset(mp, 0, sizeof(mp));
        for (int i = 1; i <= n; i ++) {
            for (int j = 1; j <= n; j ++) {
                scanf("%d", &tmp);
                if (tmp) mp[i][j] = 1;
            }
        }
        int ans = hungry();
        if (ans < n) printf("-1
");
        else {
            int cnt = 0;
            for (int i = 1; i <= n; i ++) {
                int j;
                for (j = i; j <= n; j ++) {
                    if (linker[j] == i) break;
                }
                if (j != i) {
                    cnt ++;
                    a[cnt] = i; b[cnt] = j;
                    swap(linker[i], linker[j]);
                }
            }
            printf("%d
", cnt);
            for (int i = 1; i <= cnt; i ++) printf("C %d %d
", a[i], b[i]);
        }
    }

    return 0;
}