nyoj237 Game Master's annoyance (minimum point coverage)

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The time limit of the game Master's annoyance: +Ms | Memory Limit:65535KB Difficulty:5

Describe

There is a legendary game master, in the leisure time played a small game, in the game, a n*n block-shaped area with many enemies, players can use bombs to blow up a row or a column of all enemies. He is a man who wants to play all kinds of games, so he decides to use as few bombs as possible to blow up all the enemies.

Now give you a status of the game, please help him to judge how many bombs needed at least to blow up all the enemies.

For example, the X represents the enemy

X. X
. X.

. X.

Then, he only needs to blow up the 1th and 2nd columns to blow up all the enemies, so only two bombs are needed.

Input

The

first line is an integer T that represents the number of groups of test data (0<t<=400).
The first row of each set of test data has two integer n,k, where n represents the size of the game's Square area. (N<=500,k<=10 000)
The subsequent K-line, with two integers per line, i,j denotes line I, and Column J has an enemy (both rows and columns are numbered starting with 1). (1<=i,j<=n)

Output

for each set of test data, output an integer representing the minimum number of bombs required

Sample input

13 41 11 32 23 2

Sample output

2

Source

Poj translated.

Uploaded by

Zhang Yunzun

Test instructions is to use the least amount of bombs to eliminate all enemies, which is the minimum point coverage.

The minimum point overlay can be achieved by a maximum match.

The best algorithm to find the most matches is the Hungarian algorithm

And how the composition in this question becomes a difficult place to understand. We can think of that.

Because bombs can blow up a row or a column

If there is an enemy in row J of line I, it can be imagined that I and J Point are related.

The question becomes click Open link

#include <stdio.h> #include <vector> #include <string.h>using namespace std;vector<int>map[ 505];bool vis[505];int conn[505];bool dfs (int x) {for (int i=0;i<map[x].size (); i++) {int y=map[x][i];if (!vis[y]) {Vis [Y]=true;if (conn[y]==0| | DFS (Conn[y])) {Conn[y]=x;return true;}}} return false;} int main () {int ncase;scanf ("%d", &ncase), while (ncase--) {int N,k;memset (CONN,0,SIZEOF (conn)); Memset (map,0, sizeof (map); scanf ("%d%d", &n,&k), for (int i=0;i<k;i++) {int a,b;scanf ("%d%d", &a,&b); Map[a].push _back (b);} int ans=0;for (int i=1;i<=n;i++) {memset (vis,false,sizeof (VIS)); if (Dfs (i)) ans++;} printf ("%d\n", ans);} return 0;}

nyoj237 Game Master's annoyance (minimum point coverage)