usaco 4.3.3 race3

这个题不难，就是求一个有向图的割点，还要求这个图化成无向图的割点。

我是先判断是不是有向图的割点，如果是割点再来判断是否是无向图的割点。因为我是用搜索判断是否连通的，在搜索过程中从开始点可以不通过要判断的那个点到达终点就表明这个点不是割点，在判断的过程中我有标记是否可以从开始点到达某个点，所以如果一个是有向图的割点，那么就只要判断它能不能从这个点出发能不能到达之前标记过的点，如果可以则不是无向图的割点。

 

代码：

/*ID: mnlm1991PROG: race3LANG: C++*/#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<vector>#include<algorithm>#include<string>#include<map>#include<set>#include<bitset>#include<queue>#include<iostream>using namespace std;const int MaxM = 101;const int MaxN = 51;struct ARC {int next;int v;};ARC arc[MaxM + MaxN];int v[MaxN];int arccnt;int N;void Init(int N){arccnt = 0;int i;for (i = 0; i < N; i++){arc[arccnt].next = -1;v[i] = arccnt++;}return;}void Insert(int x, int y){int f = v[x];while (arc[f].next != -1){if (arc[f].v == y){break;}f = arc[f].next;}if (arc[f].next == -1){arc[arccnt].next = -1;arc[f].next = arccnt++;arc[f].v = y;}return;}int vis[MaxN];bool DFS(int k){if (k == N - 1){return true;}int f = v[k];while (arc[f].next != -1){if (!vis[arc[f].v]){vis[arc[f].v] = 1;if (DFS(arc[f].v)){return true;}}f = arc[f].next;}return false;}bool IsCut(int x){int f = v[x];while (arc[f].next != -1){if (vis[arc[f].v] == 1){return false;}else if (!vis[arc[f].v]){vis[arc[f].v] = 2;if (!IsCut(arc[f].v)){return false;}}f = arc[f].next;}return true;}vector <int> ans1;vector <int> ans2;int main(){freopen("race3.in", "r", stdin);freopen("race3.out", "w", stdout);int x;while (scanf("%d", &x) != EOF){Init(MaxN);N = 0;while (x != -1){while (x != -2){Insert(N, x);scanf("%d", &x);}scanf("%d", &x);N++;}int i;ans1.clear();ans2.clear();for (i = 1; i < N - 1; i++){memset(vis, 0, sizeof(vis));vis[i] = 2;vis[0] = 1;if (!DFS(0)){ans1.push_back(i);if (IsCut(i)){ans2.push_back(i);}}}printf("%d", ans1.size());for (i = 0; i < ans1.size(); i++){printf(" %d", ans1[i]);}printf("/n%d", ans2.size());for (i = 0; i < ans2.size(); i++){printf(" %d", ans2[i]);}printf("/n");}return 0;}