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 the temple 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<R<1001), C(0<C<1001) representing the grid of temple (R means number of row and C means number of column of the grid) and the number of enemies N(0<N<1000001) inside the temple. After that there are N lines each of which contains 2 integers representing the position of the enemies in that temple. Each test case is followed by a new line (except the last one). Input is terminated when R=C=N=0. The size of the input file is around 1.3 MB.
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
最大匹配数等于最小覆盖数，然后还要输出对应关系，这个证明一并给出来了：http://www.cnblogs.com/ramanujan/articles/3326720.html
竟然WA几次了，原因是我用string输出，在整数到字符串转换时错了，AC代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define MAX_LL 0x7fffffffffffffff
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))
#define MAXN 1005
vector<int> g[MAXN];
bool s[MAXN], t[MAXN];
int lft[MAXN], rit[MAXN];
int r, c;
bool match(int u){
s[u]=true;
for(int i=0; i<g[u].size(); i++){
int v=g[u][i];
if(t[v]) continue;
t[v]=true;
if(lft[v]==-1 || match(lft[v])){
lft[v]=u;
rit[u]=v; //匹配点
return true;
}
}
return false;
}
void solve(){
int i, num=0;
memset(lft, -1, sizeof(lft));
memset(rit, -1, sizeof(rit));
for(i=1; i<=r; i++){
// memset(s, 0, sizeof(s));
memset(t, 0, sizeof(t));
if(match(i)) num++; //求匹配数，并用rit标记左节点集合中匹配的点
}
memset(s, 0, sizeof(s));
memset(t, 0, sizeof(t));
for(i=1; i<=r; i++) if(-1 == rit[i]) //从未匹配点出发，扩展alternating tree(匈牙利树)
match(i);
printf("%d", num);
for(i=1; i<=r; i++) if(!s[i]) //见证明
printf(" r%d", i);
for(i=1; i<=c; i++) if(t[i])
printf(" c%d", i);
printf("
");
}
int main(){
//freopen("C:\Users\Administrator\Desktop\in.txt","r",stdin);
int n;
while(scanf(" %d %d %d", &r, &c, &n)==3 && r && c && n){
int i, u, v;
for(i=1; i<=r; i++) g[i].clear();
for(i=0; i<n; i++){
scanf(" %d %d", &u, &v);
g[u].push_back(v);
}
solve();
}
return 0;
}