PKU 2186 Popular Cows

看完解题报告敲的....
没啥好说的做个模板贴出来-_-
Popular Cows

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Time Limit:1sMemory limit:32MAccepted Submit:25Total Submit:64

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Every cow's dream is to become the most popular cow in the herd. In aherd of N (1 <= N <= 10,000) cows, you are given up to M (1 <=M <= 50,000) ordered pairs of the form (A, B) that tell you that cowA thinks that cow B is popular. Since popularity is transitive, if Athinks B is popular and B thinks C is popular, then A will also thinkthat C is popular, even if this is not explicitly specified by anordered pair in the input. Your task is to compute the number of cowsthat are considered popular by every other cow.

Input

* Line 1: Two space-separated integers, N and M

* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular.

Output

* Line 1: A single integer that is the number of cows who are considered popular by every other cow.

Sample Input

3 3
1 2
2 1
2 3

Sample Output

1

Hint

Cow 3 is the only cow of high popularity.

这里把解题报告里面的一些思路贴出来 。。
容易发现，新图中唯一的出度为0的点即为所求。
因为新图不含有环，这样的点一定存在。
如果出度为0的点不唯一则无解。
.O{font-size:149%;}

•

1. #include<iostream>
2. #include<vector>
3. using namespace std;
4. #define SIZE 10050
5. typedef vector<int> VI;
6. typedef vector<VI> VII;
7. VII adj;//原图
8. VII tadj;//反向图
9. bool v[SIZE];//标记是否访问
10. int ttime[SIZE];//顺序
11. int n,m,top,scc;//scc标号
12. int id[SIZE];
13. int idtotal[SIZE];
14. int out[SIZE];//出度
15. int num;
16. void dfs1(int i)//按顺序标记点
17. {
18.  int k,temp;
19.  for(k=0;k<adj[i].size();k++)
20.  {
21.   temp=adj[i][k];//下一个点
22.   if(!v[temp])
23.   {
24.    v[temp]=true;
25.    dfs1(temp);
26.    ttime[top++]=temp;
27.   }
28.  }
29. }
30. void dfs2(int i)
31. {
32.  int k,temp;
33.  id[i]=scc;
34.  for(k=0;k<tadj[i].size();k++)
35.  {
36.   temp=tadj[i][k];
37.   if(!v[temp])
38.   {
39.    v[temp]=true;
40.    num++;
41.    dfs2(temp);  
42.   }
43.  }
44. }
45. void init()
46. {
47.  int i,x,y;
48.  top=0;
49.  adj.resize(n+1);
50.  tadj.resize(n+1);
51.  for(i=0;i<adj.size();i++)adj[i].clear();
52.  for(i=0;i<tadj.size();i++)tadj[i].clear();
53.  fill_n(out,SIZE,0);
54.  for(i=1;i<=m;i++)
55.  {
56.   scanf("%d%d",&x,&y);
57.   adj[x].push_back(y);
58.   tadj[y].push_back(x);
59.  }
60. }
61. void solve()
62. {
63.  int i;;
64.  int len,j,temp;
65.  init();
66. fill_n(v,SIZE,false);
67. for(i=1;i<=n;i++)
68.  {
69.   if(!v[i])
70.   {
71.    v[i]=true;
72.    dfs1(i);
73.    ttime[top++]=i;
74.   }
75.  }
76. fill_n(v,SIZE,false);
77. fill_n(idtotal,SIZE,0);
78.  scc=0;
79.  for(i=top-1;i>=0;i--)//逆序
80.  {
81.   if(!v[ttime[i]])
82.   {
83.    num=1;scc++;
84.    v[ttime[i]]=true;
85.    dfs2(ttime[i]);
86.    idtotal[scc]=num;//子图编号的总的顶点数
87.   }
88.  }
89.  int find_ans=0;
90.  int ans;
91.  for(i=1;i<=n;i++)
92.  {
93.   len=adj[i].size();
94.   for(j=0;j<len;j++)
95.   {
96.    temp=adj[i][j];
97.    if(id[temp]!=id[i])out[id[i]]++;
98.   }
99.  }
100.  for(i=1;i<=scc;i++)
101.  {
102.      if(!out[i])find_ans++,ans=idtotal[i];
103.  }
104. printf("%d/n",find_ans==1?ans:0);
105. }
106. int main()
107. {
108.  while(scanf("%d%d",&n,&m)!=EOF)solve();
109.  return 0;
110. }

下面是自己写的

1. #include <iostream>
2. #include <vector>
3. using namespace std;
4. typedef vector<int> VI;
5. typedef vector<VI> VII;
6. #define N 10001
7. VII mat1,mat2;
8. bool v[N];
9. int o[N],olen,n,t,out[N],ccnt[N],clen,cnt,pre[N];
10. void mark(int x)
11. {
12.     int i,tmp,tlen=mat1[x].size();
13.     for(i=0;i<tlen;i++)
14.     {
15.         tmp=mat1[x][i];
16.         if(!v[tmp]){
17.         
18.             v[tmp]=true;
19.             mark(tmp);
20.             o[olen++]=tmp;
21.         }
22.     }
23. }
24. void dfs(int x)
25. {
26.     int i,tmp,tlen=mat2[x].size();
27.     pre[x]=clen;
28.     for(i=0;i<tlen;i++){
30.         tmp=mat2[x][i];
31.         if(!v[tmp]){
32.             v[tmp]=true;
33.             cnt++;
34.             dfs(tmp);
35.         }
36.     }
37. }
38. inline void init()
39. {
40.     int i,x,y;
41.     mat1.resize(n+1);mat2.resize(n+1);
42.     for(i=olen=0;i<=n;i++)mat1[i].clear(),mat2[i].clear();
43.     for(i=0;i<t;i++)scanf("%d%d",&x,&y),mat1[x].push_back(y),mat2[y].push_back(x);
44. }
45. inline void solve()
46. {
47.     int i,j,tmp,tlen,ans;
48.     memset(v,false,sizeof(v));
49.     for(i=1;i<=n;i++){
50.         if(!v[i])
51.             {
52.                 v[i]=true;
53.                 mark(i);
54.                 o[olen++]=i;
55.             }
56.     }
57.     memset(v,false,sizeof(v));
58.     memset(ccnt,0,sizeof(ccnt));
59.     memset(out,0,sizeof(out));
60.     clen=0;
61.     for(i=olen-1;i>=0;i--){
62.         if(!v[o[i]]){
63.         cnt=1;
64.         clen++;
65.         v[o[i]]=true;
66.         dfs(o[i]);
67.         ccnt[clen]=cnt;
68.         }
69.     }
70.     for(i=1;i<=n;i++){
72.         tlen=mat1[i].size();
73.         for(j=0;j<tlen;j++){
74.             tmp=mat1[i][j];
75.             if(pre[tmp]!=pre[i])out[pre[i]]++;
76.         }
77.     }
78.     for(i=1,tmp=0;i<=clen;i++){
79.         if(out[i]==0)tmp++,ans=ccnt[i];
80.     }
81.     if(tmp==1)printf("%d/n",ans);
82.     else printf("0/n");
83. }
84. int main()
85. {
86.     while(scanf("%d%d",&n,&t)!=EOF){init();solve();}
87.     return 0;
88. }