Problem C
SAM I AM
Input: Standard Input

Output: Standard Output

The world is in great danger!! Mental's forces have returned to Earth to eradicate humankind. Our last hope to stop this great evil is Sam “Serious” Stone. Equipped with various powerful weapons, Serious Sam starts his mission to destroy the forces of evil.

After fighting two days and three nights, Sam is now in front of the temple KOPTOS where Mental's general Ugh Zan III is waiting for him. But this time, he has a serious problem. He is in shortage of ammo and a lot of enemies crawling inside the temple waiting for him. After rounding thetemple Sam finds that the temple is in rectangle shape and he has the locations of all enemies in the temple.

All of a sudden he realizes that he can kill the enemies without entering the temple using the great cannon ball which spits out a gigantic ball bigger than him killing anything it runs into and keeps on rolling until it finally explodes. But the cannonball can only shoot horizontally or vertically and all the enemies along the path of that cannon ball will be killed.

Now he wants to save as many cannon balls as possible for fighting with Mental. So, he wants to know the minimum number of cannon balls and the positions from which he can shoot the cannonballs to eliminate all enemies from outside that temple.

Input

Here, the temple is defined as a RXC grid. The first line of each test case contains 3 integers: R(0

Output

For each test case there will be one line output. First print the minimum number (m) of cannonballs needed to wipe out the enemies followed by a single space and then m positions from which he can shoot those cannonballs. For shooting horizontally print “r” followed by the row number and for vertical shooting print “c” followed by the column number. If there is more than one solution any one will do.

Sample Input Output for Sample Input

4 4 3

1 1

1 4

3 2

4 4 2

1 1

2 2

0 0 0

2 r1 r3

2 r1 r2

Problemsetter: Syed Monowar Hossain

Special Thanks: Derek Kisman

http://www.matrix67.com/blog/archives/116 文中提到從右邊未匹配點開始增廣，左邊標記的和右邊未標記的即為所求，但是也可以換一種方法，從左邊未匹配點開始增廣，左邊未標記的和右邊標記的即為所求。

#include 
#include 
#include 
#include 
using namespace std;

const int maxn = 1000+10;
int r,c,n;
struct Edge{
    int nxt,to;
}edge[maxn*maxn];
int head[maxn],from[maxn];
bool visS[maxn];
bool visT[maxn];
bool S[maxn];
int cnt;

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

void addedge(int u,int v){
    ++cnt;
    edge[cnt].nxt = head[u];
    edge[cnt].to = v;
    head[u] = cnt;
}

bool dfs(int x){
    visS[x] = 1;
    for(int i = head[x]; i != -1; i = edge[i].nxt){
        int v = edge[i].to;
        if(!visT[v]){
            visT[v] = 1;
            if(from[v]==-1||dfs(from[v])){
                from[v] = x;
                return true;
            }
        }
    }
    return false;
}

int hungary(){
    int ans = 0;
    for(int i = 1; i <= r; i++){
        for(int j = 1; j <= c; j++) visT[j] = 0;
        if(dfs(i)) ans++;
    }
    return ans;
}

int main(){

    while(cin >> r >> c >> n && r+c+n){
            init();
            for(int i = 0; i < n; i++){
                int a,b;
                scanf("%d%d",&a,&b);
                addedge(a,b);
            }
            int ans = hungary();
            cout<