HDU-2819-Swap [二分匹配][输出路径]

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题目传送门

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题意:给定一个只有0 1的矩形，问能否通过交换行列使对角线上的数全为1，如果可以输出交换的路径。

思路:首先通过匈牙利算法算出最大匹配，如果最大匹配数不等于矩阵的行数，则必然不会成功。只交换行，或者只交换列都可以交换出来。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>using namespace std;int n;bool path[110][110];bool used[110];int match[110];bool Dfs(int x){    for (int i = 1; i <= n; i++)    {        if (!used[i] && path[x][i])        {            used[i]=1;            if (!match[i] || Dfs(match[i]))            {                match[i] = x;                return true;            }        }    }    return false;}int Matching(){    int ans = 0;    memset(match,0, sizeof(match));    for (int i = 1; i <= n; i++)    {        memset(used, false, sizeof(used));        if (Dfs(i))        {            ans++;        }    }    return ans;}int main(void){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    while (~scanf("%d", &n))    {        memset(path, false, sizeof(path));        for (int i = 1; i <= n; i++)        {            for (int j = 1; j <= n; j++)            {                int x;                scanf("%d", &x);                path[i][j] = x;            }        }        int ans = Matching();        if (ans!=n)        {            printf("-1\n");            continue;        }        int L[110], R[110];        int cnt=0;        for (int i = 1; i <= n; i++)        {            if (match[i] != i)            {                for (int j = 1; j <= n; j++)                {                    if (i==match[j])                    {                        L[cnt] = i;                        R[cnt] = j;                        cnt++;                        swap(match[i], match[j]);                        break;                    }                }            }        }        printf("%d\n", cnt);        for (int i = 0; i < cnt; i++)            printf("C %d %d\n", L[i], R[i]);    }    return 0;}