hoj2819

匈牙利算法。在二分图中每行对应一个节点，每列对应一个节点，每个1对应一条连接该1所在行和列节点的边。如果最大匹配数==n则可行，否则不行。

对于可行的情况，只需要每次交换两个列节点的编号，直到所有的匹配都是行号==列号的为止。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;

#define maxn 105

int n, ptn[maxn], ans[maxn][2], tot;
bool vis[maxn];
vector< vector<int> > G(maxn);

void input()
{
    for (int i = 0; i < n; i++)
        G[i].clear();
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        {
            int a;
            scanf("%d", &a);
            if (a == 1)
                G[i].push_back(j);
        }
}

bool find(int a)
{
    for (int i = 0; i < G[a].size(); i++)
        if (!vis[G[a][i]])
        {
            if (ptn[G[a][i]] == -1)
            {
                ptn[G[a][i]] = a;
                return true;
            }
            vis[G[a][i]] = true;
            if (find(ptn[G[a][i]]))
            {
                ptn[G[a][i]] = a;
                return true;
            }
        }
    return false;
}

void work()
{
    tot = 0;
    while (1)
    {
        int a = -1;
        for (int i = 0; i < n; i++)
            if (ptn[i] != i)
            {
                a = i;
                break;
            }
        if (a == -1)
            return;
        while (ptn[a] != a)
        {
            ans[tot][0] = a;
            ans[tot][1] = ptn[a];
            tot++;
            int x = ptn[a];
            swap(ptn[a], ptn[x]);
        }
    }
}

int main()
{
    //freopen("D:\\t.txt", "r", stdin);
    while (scanf("%d", &n) != EOF && n != -1)
    {
        input();
        int ansi = 0;
        memset(ptn, -1, sizeof(ptn));
        for (int i = 0; i < n; i++)
        {
            memset(vis, 0, sizeof(vis));
            if (find(i))
                ansi++;
        }
        if (ansi < n)
        {
            printf("-1\n");
            continue;
        }
        work();
        printf("%d\n", tot);
        for (int i = 0; i < tot; i++)
            printf("C %d %d\n", ans[i][0] + 1, ans[i][1] + 1);
    }
    return 0;
}