Network of Schools
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 13325 | Accepted: 5328 |
Description
A number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintains a list of schools to which it distributes software (the “receiving schools”). Note that if B is in the distribution list of school A, then A does not necessarily appear in the list of school B 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.
Input
The first line contains an integer N: the number of schools in the network (2 <= N <= 100). The schools are identified by the first N positive integers. Each of the next N lines describes a list of receivers. The line i+1 contains the identifiers of the receivers of school i. Each list ends with a 0. An empty list contains a 0 alone in the line.
Output
Your program should write two lines to the standard output. The first line should contain one positive integer: the solution of subtask A. The second line should contain the solution of subtask B.
Sample Input
52 4 3 04 5 0001 0
Sample Output
12
下面解析转自
强连通分量缩点求入度为0的个数和出度为0的分量个数
题目大意：N(2<N<100)各学校之间有单向的网络，每一个学校得到一套软件后，能够通过单向网络向周边的学校传输，问题1：初始至少须要向多少个学校发放软件。使得网络内全部的学校终于都能得到软件。2，至少须要加入几条传输线路(边)，使随意向一个学校发放软件后，经过若干次传送，网络内全部的学校终于都能得到软件。 也就是： — 给定一个有向图，求： 1) 至少要选几个顶点。才干做到从这些顶点出发，能够到达所有顶点 2) 至少要加多少条边。才干使得从不论什么一个顶点出发。都能到达所有顶点 解题思路： — 1. 求出全部强连通分量 — 2. 每一个强连通分量缩成一点，则形成一个有向无环图DAG。 — 3. DAG上面有多少个入度为0的顶点，问题1的答案就是多少 在DAG上要加几条边，才干使得DAG变成强连通的，问题2的答案就是多少 加边的方法： 要为每一个入度为0的点加入入边，为每一个出度为0的点加入出边 假定有 n 个入度为0的点，m个出度为0的点。怎样加边？ 把全部入度为0的点编号 0,1,2,3,4 ....N -1 每次为一个编号为i的入度0点可达的出度0点，加入一条出边，连到编号为(i+1)%N 的那个出度0点, 这须要加n条边 若 m <= n，则 加了这n条边后。已经没有入度0点。则问题解决。一共加了n条边 若 m > n。则还有m-n个入度0点，则从这些点以外任取一点，和这些点都连上边。就可以，这还需加m-n条边。 所以，max(m,n)就是第二个问题的解 此外：当仅仅有一个强连通分支的时候，就是缩点后仅仅有一个点，尽管入度出度为0的都有一个，可是实际上不须要添加清单的项了，所以答案是1。0；#include#include #include #include #define maxn 100 + 100#define maxm 110 * 100using namespace std;int n, m;struct node{ int u, v, next;};node edge[maxm];int head[maxn], cnt;int low[maxn], dfn[maxn];int dfs_clock;int Stack[maxn], top;bool Instack[maxn];int Belong[maxn];int scc_clock;int in[maxn], out[maxn];int num[maxn];void init(){ cnt = 0; memset(head, -1, sizeof(head));}void addedge(int u, int v){ edge[cnt] = {u, v, head[u]}; head[u] = cnt++;}void getmap(){ for(int i = 1; i <= n; ++i){ int v; while(scanf("%d", &v),v){ addedge(i, v); } }}void Tarjan(int u, int per){ int v; low[u] = dfn[u] = ++dfs_clock; Stack[top++] = u; Instack[u] = true; for(int i = head[u]; i != -1; i = edge[i].next){ int v = edge[i].v; if(!dfn[v]){ Tarjan(v, u); low[u] = min(low[u], low[v]); } else if(Instack[v]) low[u] = min(low[u], dfn[v]); } if(dfn[u] == low[u]){ scc_clock++; do{ v = Stack[--top]; Instack[v] = false; Belong[v] = scc_clock; } while( v != u); }}void suodian(){ for(int i = 1; i <= scc_clock; ++i){ out[i] = 0; in[i] = 0; } for(int i = 0; i < cnt; ++i){ int u = Belong[edge[i].u]; int v = Belong[edge[i].v]; if(u != v){ out[u]++; in[v]++; } }}void find(){ memset(low, 0, sizeof(low)); memset(dfn, 0, sizeof(dfn)); memset(Belong, 0, sizeof(Belong)); memset(Stack, 0, sizeof(Stack)); memset(Instack, false, sizeof(false)); dfs_clock = scc_clock = top = 0; for(int i = 1; i <= n ; ++i){ if(!dfn[i]) Tarjan(i, i); }}void solve(){ if(scc_clock == 1){ printf("1\n0\n"); return ; } int numin, numout; numin = numout = 0; for(int i = 1; i <= scc_clock; ++i){ if(in[i] == 0) numin++; if(out[i] == 0) numout++; } printf("%d\n%d\n", numin, max(numout, numin));}int main (){ while(scanf("%d", &n) != EOF){ init(); getmap(); find(); suodian(); solve(); } return 0;}
