HDU 2819 ——Swap——————【最大匹配、利用linker数组、邻接表方式】

 Swap

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2819

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

 

 

题目大意：给你一个n*n的矩阵，里面只有0、1，问你需要几步可以让主对角线上全是1。如果不能实现，输出-1.

 

解题思路：如果想一想，画一画，可以发现，如果有行或者列全是0或1的话，那么结果肯定是-1。否则一定有解。按照常规的行列分部，在有1的行列进行连边。我们跑一遍最大匹配。如果最大匹配小于n，说明无解。否则利用匹配数组linker。这里linker[i]表示第i列跟第linker[i]行交点有数字1。那么我们枚举第i列。如果linker[i] != i，说明这个位置需要调换，然后暴力枚举linker[j] == i的j，交换linker[i],linker[j]，同时记录路径即可。

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn = 1100;
struct Edge{
    int from, to, dist, next;
    Edge(){}
    Edge(int _from,int _to,int _next):from(_from),to(_to),next(_next){}
}edges[maxn*maxn*3];    //direction
struct Swap{
    int x,y;
}swaps[maxn*maxn];
int tot , head[maxn];
int linker[3*maxn], used[3*maxn], c[maxn];
void init(){
    tot = 0;
    memset(head,-1,sizeof(head));
}
void AddEdge(int _u,int _v){
    edges[tot] = Edge(_u,_v,head[_u]);
    head[_u] = tot++;
}
bool dfs(int u,int _n){
    for(int e = head[u]; e != -1; e = edges[e].next){
        int v = edges[e].to;
        if(!used[v]){
            used[v] = u;
            if(linker[v] == -1 || dfs(linker[v],_n)){
                linker[v] = u;
                return true;
            }
        }
    }
    return false;
}
int hungary(int p, int n){
    int ret = 0;
    memset(linker,-1,sizeof(linker));
    for(int i = 1; i <= p; i++){
        memset(used,0,sizeof(used));
        if(dfs(i,n))
            ret++;
    }
    return ret;
}
int main(){
    int n, m, T, p, k, cas = 0;
    while(scanf("%d",&n)!=EOF){
        int a,b;
        init();
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){
                scanf("%d",&a);
                if(a == 1){
                    AddEdge(i,j);
                }
            }
        }
        int ans = hungary(n,m);
        if(ans < n ){
            puts("-1"); continue;
        }
        int num = 0;
        for(int i = 1; i <= n; i++){
            if(linker[i] != i){
                for(int j = i+1; j <= n; j++){
                    if(linker[j] == i){
                        swap(linker[i],linker[j]);
                        num++;
                        swaps[num].x = i; swaps[num].y = j;
                    }
                }
            }
        }
        printf("%d
",num);
        for(int i = 1; i <= num; i++){
            printf("C %d %d
",swaps[i].x,swaps[i].y);
        }
    }
    return 0;
}