Network of Schools POJ - 1236 tarjan强连通分量缩点
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Network of Schools POJ - 1236 tarjan强连通分量缩点
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You are to write a program that computes the minimal number of schools that must receive a copy of the new software in order for the software to reach all schools in the network according to the agreement (Subtask A). As a further task, we want to ensure that by sending the copy of new software to an arbitrary school, this software will reach all schools in the network. To achieve this goal we may have to extend the lists of receivers by new members. Compute the minimal number of extensions that have to be made so that whatever school we send the new software to, it will reach all other schools (Subtask B). One extension means introducing one new member into the list of receivers of one school.?
題解：
題目第一句話說所有的電腦都是相互連接的，所以不用擔(dān)心出現(xiàn)森林的情況。
第一步用tarjan進(jìn)行縮點(diǎn)，重新構(gòu)圖得到一棵樹，然后統(tǒng)計(jì)樹上各點(diǎn)的入度以及出度。
記in為入讀為0的點(diǎn)，記out為出度為0的點(diǎn)。
那么我們很顯然的要給所有入度為0的點(diǎn)全都分配一套軟件，因?yàn)樗麄儫o法從別的地方得到軟件。
所以第一問的答案就是in。
考慮第二問，如果out > in 的話，我們從所有的出度為0的點(diǎn)開始連一條邊到入度為0的點(diǎn)去，這樣的話，入讀為0的點(diǎn)一定會(huì)被覆蓋，而且有的點(diǎn)甚至有多余1條入邊，這無所謂。所以我們只需要添加out條邊就可以了。
如果in > out的話，同理。
因此我們添加的邊的條數(shù)應(yīng)該為max(in,out)
代碼：
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <set> using namespace std; const int MAXN = 107; int head[MAXN]; int cnt; int DFN[MAXN]; int LOW[MAXN]; int visit[MAXN]; int scc[MAXN]; int belong[MAXN]; int stk[MAXN];int sp; int index; int sccnum; struct edge{ int v; int next; int cost; }Es[10007]; void init(){ sp = sccnum = index = cnt = 0; memset(head,-1,sizeof(head)); memset(visit,0,sizeof(visit)); memset(belong,0,sizeof(belong)); } inline void add_edge(int i,int j,int cost){ Es[cnt].v = j; Es[cnt].cost = cost; Es[cnt].next = head[i]; head[i] = cnt++; } void tarjan(int u){DFN[u] = LOW[u] = ++index;visit[u] = 1;stk[sp++] = u; for(int e = head[u];e != -1;e = Es[e].next){int v = Es[e].v;if(!DFN[v]){tarjan(v);LOW[u] = min(LOW[u],LOW[v]);}else if(visit[v]){LOW[u] = min(LOW[u],DFN[v]);}}if(DFN[u] <= LOW[u]){//sccnum++;int top;do{top = stk[--sp];belong[top] = sccnum;visit[top] = 0;}while(top != u);} }int main() {init();int N;scanf("%d",&N);for(int i = 1;i <= N;i++){int num;while(scanf("%d",&num) && num){add_edge(i,num,1);}}for(int i = 1;i <= N;i++){if(!DFN[i])tarjan(i);}set<int> st1,st2;for(int i = 1;i <= N;i++){for(int e = head[i];e != -1;e = Es[e].next){int v = Es[e].v;if(belong[i] == belong[v]) continue;st1.insert(belong[i]);st2.insert(belong[v]);}}if(sccnum == 1){puts("1\n0");return 0;}cout<<sccnum - st2.size()<<endl;cout<<max(sccnum - st1.size(),sccnum - st2.size())<<endl; }
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