UVA - 11419 SAM I AM

Description

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

 

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 00

2 r1 r3 2 r1 r2

---

Problemsetter: SyedMonowarHossain

Special Thanks: Derek Kisman

题意：求最少的打掉所有的障碍物

思路：二分图最小的点覆盖，证明就自己找吧

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int MAXN = 1010;

vector G[MAXN];
int visl[MAXN], visr[MAXN];
int map[MAXN][MAXN];
int arrl[MAXN], arrr[MAXN];
int n, m, k;

void init() {
	memset(arrl, -1, sizeof(arrl));
	memset(arrr, -1, sizeof(arrr));
	memset(visr, 0, sizeof(visr));
	memset(visl, 0, sizeof(visl));
	memset(map, 0, sizeof(map));
	for (int i = 0; i < MAXN; i++)
		G[i].clear();
}

int dfs(int u) {
	visl[u] = 1;
	for (int i = 0; i < G[u].size(); i++) {
		int v = G[u][i];		
		if (!visr[v]) {
			visr[v] = 1;
			if (arrl[v] == -1 || dfs(arrl[v])) {
				arrl[v] = u;
				arrr[u] = v;
				return 1;
			}
		}
	}	
	return 0;
}

int main() {
	while (scanf("%d%d%d", &n, &m, &k) != EOF && n+m+k) {
		init();
		for (int i = 0; i < k; i++) {
			int x, y;
			scanf("%d%d", &x, &y);
			G[x].push_back(y);
		}
		int ans = 0;
		for (int i = 1; i <= n; i++) {
			memset(visr, 0, sizeof(visr));
			if (dfs(i))
				ans++;
		}
		printf("%d", ans);
		memset(visl, 0, sizeof(visl));
		memset(visr, 0, sizeof(visr));
		for (int i = 1; i <= n; i++) 
			if (arrr[i] == -1)
				dfs(i);
		for (int i = 1; i <= n; i++)
			if (!visl[i])
				printf(" r%d", i);
		for (int i = 1; i <= m; i++)
			if (visr[i])
				printf(" c%d", i);
		printf("\n");
	}	
	return 0;
}