POJ 1236 —— Network of Schools

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原题：http://poj.org/problem?id=1236

题意：有n个点，下面n行给出与第i个点相连的结点；在某个点放置软件，那么与该点相连的点也可以得到该软件，问最少要在几个点放置软件；第二问是至少要添加几条边使得该图强连通；

思路：通过强连通求得DAG图，第一问就是求新图中入度 = 0 的点的个数；第二问就是求新图中max ( 入度 = 0 的点个数，出度 = 0 的点个数 )；

注意：如果强连通分量只有一个，那么输出（1\n0\n）；

#include<cstdio>#include<cstring>#include<string>#include<algorithm>using namespace std;const int maxn = 110;int n;int head[maxn], edgenum;int Stack[maxn], Belong[maxn], DFN[maxn], Low[maxn];int Time, taj, top;bool Instack[maxn];int in[maxn], out[maxn];struct Edge{int from, to;int next;}edge[maxn*maxn];void Tarjan(int u){DFN[u] = Low[u] = ++Time;Stack[top++] = u;Instack[u] = true;for(int i = head[u];i != -1;i = edge[i].next){int v = edge[i].to;if(DFN[v] == -1){Tarjan(v);Low[u] = min(Low[u], Low[v]);}else{if(Instack[v] && Low[u] > DFN[v])Low[u] = DFN[v];}}if(DFN[u] == Low[u]){taj++;while(1){int now = Stack[--top];Belong[now] = taj;Instack[now] = false;if(now == u)break;}}}void add(int u, int v){edge[edgenum].from = u;edge[edgenum].to = v;edge[edgenum].next = head[u];head[u] = edgenum++;}void init(){memset(head, -1, sizeof head);edgenum = 0;memset(DFN, -1, sizeof DFN);memset(Instack, false, sizeof Instack);Time = taj = top = 0;}int main(){while(~scanf("%d", &n)){init();for(int i = 1;i<=n;i++){int v;while(scanf("%d", &v) && v){add(i, v);}}for(int i = 1;i<=n;i++){if(DFN[i] == -1)Tarjan(i);}memset(in, 0, sizeof in);memset(out, 0, sizeof out);for(int i = 0;i<edgenum;i++){int u = Belong[edge[i].from];int v = Belong[edge[i].to];if(u != v){out[u]++;in[v]++;}}if(taj == 1){printf("1\n0\n");continue;}int incnt = 0, outcnt = 0;for(int i = 1;i<=taj;i++){if(in[i] == 0)incnt++;if(out[i] == 0)outcnt++;}printf("%d\n%d\n", incnt, max(incnt, outcnt));}return 0;}

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