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4 0 3 2 1 2 1 3
4 2分析:题意:有1~N代表N个结论,首先已给出m个关系,a b 代表a能证明b,为至少需要添加几条边能是N条结论相互两两能够证明,很明显就是至少添加几条边使构成一个强联通。先缩点,在求出入度为0,和出度为0的个数,取其最大这便为答案。
- #include<stdio.h>
- #include<string.h>
- //#include<algorithm>
- //#include<iostream>
- #include<stack>
- #define Max(a,b) a>b?a:b
- #define maxh 220000+10
- #define maxn 550000+10
- using namespace std;
- typedef struct{
- int to,next;
- }node;
- stack<int>S;
- node E1[maxn],E2[maxn];
- int head1[maxh],head2[maxh],mark[maxh],cnt1,cnt2;
- int belong[maxh],num[maxh][2],in[maxh],out[maxh],count;
- void init(int n){
- memset(head1,-1,sizeof(head1));
- memset(head2,-1,sizeof(head2));
- cnt1=0;
- cnt2=0;
- }
- void add(int a,int b){
- E1[cnt1].to=b,E1[cnt1].next=head1[a],head1[a]=cnt1++;
- E2[cnt2].to=a,E2[cnt2].next=head2[b],head2[b]=cnt2++;
- }
- void dfs1(int x){
- mark[x]=1;
- for(int i=head1[x];i+1;i=E1[i].next){
- int to=E1[i].to;
- if(mark[to])
- continue;
- dfs1(to);
- }
- S.push(x);
- }
- void dfs2(int x){
- mark[x]=1;
- belong[x]=count;
- for(int i=head2[x];i+1;i=E2[i].next){
- int to=E2[i].to;
- if(mark[to])
- continue;
- dfs2(to);
- }
- }
- int main(){
- int n,m,a,b,sumin,sumout;
- while(~scanf("%d%d",&n,&m)){
- init(n);
- for(int i=0;i<m;i++){
- scanf("%d%d",&a,&b);
- num[i][0]=a;
- num[i][1]=b;
- add(a,b);
- }
- memset(mark,0,sizeof(mark));
- while(!S.empty())
- S.pop();
- for(int i=1;i<=n;i++){
- if(!mark[i]){
- dfs1(i);
- }
- }
- memset(mark,0,sizeof(mark));
- count=0;
- while(!S.empty()){
- int s=S.top();
- S.pop();
- if(!mark[s]){
- count++;
- dfs2(s);
- }
- }
- if(count==1){
- printf("0\n");
- continue;
- }
- memset(in,0,sizeof(in));
- memset(out,0,sizeof(out));
- sumin=sumout=0;
- for(int i=0;i<m;i++){
- if(belong[num[i][0]]!=belong[num[i][1]])
- in[belong[num[i][0]]]=out[belong[num[i][1]]]=1;
- }
- for(int i=1;i<=count;i++){
- if(!in[i]){
- sumin++;
- }
- if(!out[i]){
- sumout++;
- }
- }
- printf("%d\n",Max(sumin,sumout));
- }
- return 0;
- }
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原文地址:http://blog.csdn.net/letterwuyu/article/details/43453297