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  • hdu2767 Proving Equivalences

    先求一发 scc 然后缩点,然后根据入度出度为 (0) 的点的个数乱搞。

    #include <iostream>
    #include <cstring>
    #include <cstdio>
    using namespace std;
    int T, n, m, hea[20005], cnt, uu, vv, bel[20005], scc, din, sta[20005], idx;
    int dfn[20005], loo[20005], ind[20005], oud[20005], cnt1, cnt2;
    bool ins[20005];
    struct Edge{
    	int too, nxt;
    }edge[100005];
    void add_edge(int fro, int too){
    	edge[++cnt].nxt = hea[fro];
    	edge[cnt].too = too;
    	hea[fro] = cnt;
    }
    void tarjan(int x){
    	dfn[x] = loo[x] = ++idx;
    	ins[x] = true;
    	sta[++din] = x;
    	for(int i=hea[x]; i; i=edge[i].nxt){
    		int t=edge[i].too;
    		if(!dfn[t]){
    			tarjan(t);
    			loo[x] = min(loo[x], loo[t]);
    		}
    		else if(ins[t])	loo[x] = min(loo[x], dfn[t]);
    	}
    	if(dfn[x]==loo[x]){
    		scc++;
    		int j;
    		do{
    			j = sta[din--];
    			ins[j] = false;
    			bel[j] = scc;
    		}while(dfn[j]!=loo[j]);
    	}
    }
    int main(){
    	cin>>T;
    	while(T--){
    		memset(hea, 0, sizeof(hea));
    		memset(dfn, 0, sizeof(dfn));
    		memset(loo, 0, sizeof(loo));
    		memset(ind, 0, sizeof(ind));
    		memset(oud, 0, sizeof(oud));
    		cnt = scc = din = idx = cnt1 = cnt2 = 0;
    		scanf("%d %d", &n, &m);
    		for(int i=1; i<=m; i++){
    			scanf("%d %d", &uu, &vv);
    			add_edge(uu, vv);
    		}
    		for(int i=1; i<=n; i++)
    			if(!dfn[i])
    				tarjan(i);
    		for(int i=1; i<=n; i++)
    			for(int j=hea[i]; j; j=edge[j].nxt){
    				int t=edge[j].too;
    				if(bel[i]!=bel[t]){
    					ind[bel[t]]++;
    					oud[bel[i]]++;
    				}
    			}
    		for(int i=1; i<=scc; i++){
    			if(!ind[i])	cnt1++;
    			if(!oud[i])	cnt2++;
    		}
    		if(scc==1)	printf("0
    ");
    		else
    			printf("%d
    ", max(cnt1, cnt2));
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/poorpool/p/8584194.html
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