Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 13800 | Accepted: 5504 |
Description
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
Output
Sample Input
5 2 4 3 0 4 5 0 0 0 1 0
Sample Output
1 2
题意:向学校发软件,1、问最少向多少个学校发软件就可以间接让所有学校都得到软件 2、问最少添加多少条边就可以使 任意向一所学校发软件 就可以让其余所有学校都收到软件
题解:求出强连通分支后缩点,求出所有入度和出度问0的点的个数insum和outsum,则答案为insum和max(insum,outsum)
输入:第一行是一个n代表接下来有n行,每行输入小于n个数以0结尾,代表这个点与第i行(当前行数)的值i点相连
#include<stdio.h> #include<string.h> #include<stack> #include<vector> #include<algorithm> #define MAX 1100 #define MAXM 2001000 #define INF 0x7ffffff using namespace std; int low[MAX],dfn[MAX]; int head[MAX],ans; int instack[MAX]; int dfsclock,scccnt; int in[MAX],out[MAX]; int sccno[MAX]; stack<int>s; vector<int>newmap[MAX]; vector<int>scc[MAX]; struct node { int beg,end,next; }edge[MAXM]; void init() { ans=0; memset(head,-1,sizeof(head)); } void add(int u,int v) { edge[ans].beg=u; edge[ans].end=v; edge[ans].next=head[u]; head[u]=ans++; } void tarjan(int u) { int v,i; s.push(u); instack[u]=1; low[u]=dfn[u]=++dfsclock; for(i=head[u];i!=-1;i=edge[i].next) { v=edge[i].end; if(!dfn[v]) { tarjan(v); low[u]=min(low[u],low[v]); } else if(instack[v]) low[u]=min(low[u],dfn[v]); } if(low[u]==dfn[u]) { scccnt++; while(1) { v=s.top(); s.pop(); instack[v]=0; sccno[v]=scccnt; if(v==u) break; } } } void find(int l,int r) { dfsclock=scccnt=0; memset(low,0,sizeof(low)); memset(dfn,0,sizeof(dfn)); memset(instack,0,sizeof(instack)); memset(sccno,0,sizeof(sccno)); for(int i=l;i<=r;i++) { if(!dfn[i]) tarjan(i); } } void suodian() { int u,v; memset(in,0,sizeof(in)); memset(out,0,sizeof(out)); for(int i=0;i<ans;i++) { u=sccno[edge[i].beg]; v=sccno[edge[i].end]; if(v!=u) { newmap[u].push_back(v); in[v]++; out[u]++; } } } void solve() { int i,j; int sum,insum,outsum; insum=outsum=0; if(scccnt==1) { printf("1\n0\n"); return ; } for(i=1;i<=scccnt;i++) { if(!in[i]) insum++; if(!out[i]) outsum++; } sum=max(insum,outsum); printf("%d\n%d\n",insum,sum); } int main() { int n,m,j,i,a; while(scanf("%d",&n)!=EOF) { init(); for(i=1;i<=n;i++) { while(scanf("%d",&a),a) add(i,a); } find(1,n); suodian(); solve(); } return 0; }