HDU2819(KB10-E 二分图最大匹配)

Swap

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3800    Accepted Submission(s): 1401
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

 

按行交换。对行进行1-n编号，对列也进行编号。

若第i行的第j1、j2……列有1，则节点i与j1、j2……连边。

二分图跑最大匹配，为完美匹配方可。

matching数组保存每个点的匹配，模拟一下交换输出答案。

//2017-08-26
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 100000;
const int M = 5000000;
int head[N], tot;
struct Edge{
    int to, next;
}edge[M];

void init(){
    tot = 0;
    memset(head, -1, sizeof(head));
}

void add_edge(int u, int v){
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;

    edge[tot].to = u;
    edge[tot].next = head[v];
    head[v] = tot++;
}

int n;
int matching[N];
int check[N];
bool dfs(int u){
    for(int i =  head[u]; i != -1; i = edge[i].next){
        int v = edge[i].to;
        if(!check[v]){//要求不在交替路
            check[v] = 1;//放入交替路
            if(matching[v] == -1 || dfs(matching[v])){
                //如果是未匹配点，说明交替路为增广路，则交换路径，并返回成功
                matching[u] = v;
                matching[v] = u;
                return true;
            }
        }
    }
    return false;//不存在增广路
}

//hungarian: 二分图最大匹配匈牙利算法
//input: null
//output: ans 最大匹配数
int hungarian(){
    int ans = 0;
    memset(matching, -1, sizeof(matching));
    for(int u = 1; u <= n; u++){
        if(matching[u] == -1){
            memset(check, 0, sizeof(check));
            if(dfs(u))
              ans++;
        }
    }
    return ans;
}

const int MAXID = 100;
int a[N], b[N], cnt;

int main()
{
    std::ios::sync_with_stdio(false);
    //freopen("inputE.txt", "r", stdin);
    while(cin>>n){
        init();
        int v;
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                cin>>v;
                if(v){
                    add_edge(i, j+MAXID);
                }
            }
        }
        int match = hungarian();
        if(match != n)cout<<-1<<endl;
        else{
            for(int i = 1; i <= n; i++)
                  matching[i]-=100;
            cnt = 0;
            for(int i = 1; i <= n; i++){
                for(int j = 1; j <= n; j++){
                    if(i == j)continue;
                    if(matching[j] == i){
                        swap(matching[i], matching[j]);
                        a[cnt] = i;
                        b[cnt++] = j;
                    }
                }
            }
            cout<<cnt<<endl;
            for(int i = 0; i < cnt; i++)
                  cout<<"R "<<a[i]<<" "<<b[i]<<endl;
        }
    }

    return 0;
}