LeetCode 907 minimize-malware-spread(连通分量)

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坑点:

1. 不是删除一个顶点，是将一个顶点去除颜色

题目链接

minimize-malware-spread

分析

了解上面的坑点之后就简单多了

只有当且仅当一个 initial 点在连通分量中，我们擦除它的颜色才有收益

code

import java.util.*;
class Solution {
    public int minMalwareSpread(int[][] graph, int[] initial) {
    	Arrays.sort(initial);
    	int n = graph.length;
    	int[] comp = new int[n];
    	int[] candidate = new int[n+1];
    	boolean[] isVal = new boolean[n];
    	for(int i=0 ; i<initial.length ; ++i)isVal[initial[i]] = true;
    	int C = 0;
    	for(int i=0 ; i<graph.length ; ++i){
    		Queue<Integer> Q = new LinkedList<>();
    		if(comp[i]==0)dfs(i, ++C, comp, graph,Q,isVal);
    		if(Q.size() == 1){
    			candidate[C] = Q.remove()+1;
    		}
    	}
    	
    	int[] cnt = new int[C+1];
    	for(int i=0 ; i<n ; ++i)cnt[comp[i]]++;
    	int val = 0,ans = 0;
    	for(int i=1 ; i <=C ; ++i)
    		if(candidate[i]!=0){
    			if(cnt[i] > val|| (cnt[i] == val && ans > candidate[i])){
    				ans = candidate[i];
    				val = cnt[i];
    			}
    		}
    	
    	if(ans !=0)return ans -1;
    	else return initial[0];
    		
    }
    private void dfs(int u, int col,int[] comp,int[][] g, Collection<Integer> black,boolean[] isVal) {
		comp[u] = col;
		if(isVal[u])black.add(u);
		for(int i=0 ; i< comp.length ; ++i)
			if(g[u][i]==1 &&comp[i]==0)dfs(i, col, comp,g,black,isVal);
	}
}