填边判强连通 -- Strongly connected HDU - 4635

Strongly connected HDU - 4635

思路：

code：
莫名bug，找了一天心累了，懒得找了

#include 
#include 
#include 
using namespace std;
const int maxn = 1e5 + 5;

struct node{
	int u, v, next;
}g[maxn];
int head[maxn];
int dfn[maxn], low[maxn], stack[maxn], cot, cnt, opt, sum;
bool vis[maxn];
int index[maxn], in[maxn], out[maxn], totol[maxn];

void add(int u, int v){
	g[++cnt].u = u;
	g[cnt].v = v;
	g[cnt].next = head[u];
	head[u] = cnt;
}

void init(int n){
	cot = cnt = opt = sum = 0;
	for(int i = 1; i <= n; i++){
		dfn[i] = low[i] = 0;
		vis[i] = false;
		head[i] = 0;
	    index[i] = 0;
	    in[i] = out[i] = totol[i] = 0;     
	}
	
}

void targan(int u) {
	dfn[u] = low[u] = ++cot;
    stack[opt++] = u;
    vis[u] = true;
	for(int i = head[u]; i != 0; i = g[i].next){
		int v = g[i].v;
		if(!dfn[v]) {
			targan(v);
			low[u] = min(low[u], low[v]);
		}
		else if(vis[i]) low[u] = min(low[u], dfn[v]);
	}
	
	if(dfn[u] == low[u]){
		int res;
		sum++;
		do{
			res = stack[--opt];
			vis[res] = false;
			index[res] = sum;
			totol[sum]++;    //记录强联通分量的节点数 
		} while(res != u);
	}
	
}


int main(){
	int t, n, m, u, v;
	int ans;

    scanf("%d", &t);
	for(int ca = 1; ca <= t; ca++){
		scanf("%d%d", &n, &m);
		
		init(n);
		
		for(int i = 0; i < m; i++){
            scanf("%d%d", &u, &v);
			add(u, v);
		}
		
		for(int i = 1; i <= n; i++) 
		   if(!dfn[i]) targan(i);
		   
		for(int i = 1; i <= cnt; i++) {
			int pre = g[i].u;
			int cur = g[i].v;
			if(index[pre] == index[cur]) continue;
			out[index[pre]]++;
			in[index[cur]]++;
		}
       
		
				
		int mix = 1e9;
	    for(int i = 1; i <= sum; i++) 	    
	    	if(!in[i] || !out[i]) mix = min(mix, totol[i]);
		
	   
	   long long ans;
	   if(sum == 1) ans = -1;
	   else ans = (long long)n * (n - 1) - (long long)m - (long long)mix * (n - mix);
	   printf("Case %d: %lld\n", ca, ans);

		
	}

	
	     
}

dalao code：

#include 
#include 
#include 
typedef long long ll;
using namespace std;
const int N = 1e5 + 5, M = 1e5 + 5;
struct E { int v, next;} e[M];
int t, n, m, u, v, len, h[N], minv, scc_cnt, top, num, id[N], scc[N], ind[N], outd[N], dfn[N], low[N], stack[N];
bool in_st[N];  
void add(int u, int v) {e[++len].v = v; e[len].next = h[u]; h[u] = len;}
void tarjan(int u) {
	dfn[u] = low[u] = ++num;
	stack[++top] = u; in_st[u] = true;
	for (int j = h[u]; j ; j = e[j].next) {
		int v = e[j].v;
		if (!dfn[v]) {
			tarjan(v);
			low[u] = min(low[u], low[v]);
		} else if (in_st[v]) low[u] = min(low[u], dfn[v]);
	} 
	if (dfn[u] == low[u]) {
		int v; scc_cnt++;
		do {
			v = stack[top--]; in_st[v] = false;
			id[v] = scc_cnt; scc[scc_cnt]++;
		} while (u != v);
	}
}
int main() {  
	scanf("%d", &t); int cas = 1;
	while (t--) {
		memset(h, 0, sizeof(h)); len = num = top = scc_cnt = 0; minv = 1e9; 
		memset(in_st, false, sizeof(in_st)); 
		memset(dfn, 0, sizeof(dfn));
		memset(ind, 0, sizeof(ind));
		memset(outd, 0, sizeof(outd));
		memset(id, 0, sizeof(id));
		memset(scc, 0, sizeof(scc));
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= m; i++) {
			scanf("%d%d", &u, &v);
			add(u, v);
		}
		for (int i = 1; i <= n; i++) if (!dfn[i]) tarjan(i);
		for (int u = 1; u <= n; u++) {
			for (int j = h[u]; j; j = e[j].next) {
				int v = e[j].v;
				if (id[v] == id[u]) continue;
				ind[id[v]]++, outd[id[u]]++;
			}
		}
		//求出度为0 或者 入度为0的SCC的最少点数 
		for (int i = 1; i <= scc_cnt; i++) {
			if (!ind[i] || !outd[i]) minv = min(minv, scc[i]);
		}
		if (scc_cnt == 1) printf("Case %d: -1\n", cas++);
		else printf("Case %d: %lld\n", cas++, (ll)n * (n - 1) - (ll)m - (ll)minv * (n - minv));
	}
	return 0; 
}