Problem Description
This year is the 60th anniversary of NJUST, and to make the celebration more colorful, Tom200 is going to invite distinguished alumnus back to visit and take photos. After carefully planning, Tom200 announced his activity plan, one that contains two characters: 1. Whether the effect of the event are good or bad has nothing to do with the number of people join in. 2. The more people joining in one activity know each other, the more interesting the activity will be. Therefore, the best state is that, everyone knows each other. The event appeals to a great number of alumnus, and Tom200 finds that they may not know each other or may just unilaterally recognize others. To improve the activities effects, Tom200 has to divide all those who signed up into groups to take part in the activity at different time. As we know, one‘s energy is limited, and Tom200 can hold activity twice. Tom200 already knows the relationship of each two person, but he cannot divide them because the number is too large. Now Tom200 turns to you for help. Given the information, can you tell if it is possible to complete the dividing mission to make the two activity in best state.
Input
The input contains several test cases, terminated by EOF. Each case starts with a positive integer n (2<=n<=100), which means the number of people joining in the event. N lines follow. The i-th line contains some integers which are the id of students that the i-th student knows, terminated by 0. And the id starts from 1.
Output
If divided successfully, please output "YES" in a line, else output "NO".
Sample Input
3
3 0
1 0
1 2 0
Sample Output
YES
Source
2013 ACM/ICPC Asia Regional Nanjing Online
怒贴两种方法的代码，以表示我的愤怒，这么简单的题目都想的那么复杂
第一种是dfs染色法
1 #pragma comment(linker, "/STACK:1024000000,1024000000")
2 #include<iostream>
3 #include<cstdio>
4 #include<cstring>
5 #include<cmath>
6 #include<math.h>
7 #include<algorithm>
8 #include<queue>
9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 using namespace std;
15 #define ll long long
16 #define eps 1e-10
17 #define MOD 1000000007
18 #define N 106
19 #define inf 1e12
20 int n;
21 vector<int>v[N];
22 int color[N];
23 int mp[N][N];
24 bool dfs(int u,int c){
25 color[u]=c;
26 for(int i=0;i<v[u].size();i++){
27 int num=v[u][i];
28 if(color[num]!=-1){
29 if(color[num]==c){
30 return false;
31 }
32 continue;
33 }
34 if(!dfs(num,!c)) return false;
35 }
36 return true;
37 }
38 int main()
39 {
40 while(scanf("%d",&n)==1){
41 for(int i=0;i<N;i++){
42 v[i].clear();
43 }
44 memset(mp,0,sizeof(mp));
45 for(int i=1;i<=n;i++){
46 int x;
47 scanf("%d",&x);
48 while(x!=0){
49 //v[i].push_back(x);
50 //v[x].push_back(i);
51 mp[i][x]=1;
52 scanf("%d",&x);
53 }
54 }
55
56 for(int i=1;i<=n;i++){
57 for(int j=i+1;j<=n;j++){
58 if(mp[i][j]==0 || mp[j][i]==0){
59 v[i].push_back(j);
60 v[j].push_back(i);
61 }
62 }
63 }
64
65 memset(color,-1,sizeof(color));
66 int flag=1;
67 for(int i=1;i<=n;i++){
68 if(color[i]==-1 && !dfs(i,0)){
69 flag=0;
70 break;
71 }
72 }
73 if(flag){
74 printf("YES\n");
75 }
76 else{
77 printf("NO\n");
78 }
79 }
80 return 0;
81 }
第二种是2-sat，其实本质上和上一种是一样的
1 #pragma comment(linker, "/STACK:1024000000,1024000000")
2 #include<iostream>
3 #include<cstdio>
4 #include<cstring>
5 #include<cmath>
6 #include<math.h>
7 #include<algorithm>
8 #include<queue>
9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 #include <stack>
15 using namespace std;
16 #define PI acos(-1.0)
17 #define max(a,b) (a) > (b) ? (a) : (b)
18 #define min(a,b) (a) < (b) ? (a) : (b)
19 #define ll long long
20 #define eps 1e-10
21 #define MOD 1000000007
22 #define N 106
23 #define inf 1e12
24 int n,m;
25
26 int mp[N][N];
27 int tot;
28 int head[N];
29 int vis[N];
30 int tt;
31 int scc;
32 stack<int>s;
33 int dfn[N],low[N];
34 int col[N];
35 struct Node
36 {
37 int from;
38 int to;
39 int next;
40 }edge[N*N];
41 void init()
42 {
43 tot=0;
44 scc=0;
45 tt=0;
46 memset(head,-1,sizeof(head));
47 memset(dfn,-1,sizeof(dfn));
48 memset(low,0,sizeof(low));
49 memset(vis,0,sizeof(vis));
50 memset(col,0,sizeof(col));
51 }
52 void add(int s,int u)//邻接矩阵函数
53 {
54 edge[tot].from=s;
55 edge[tot].to=u;
56 edge[tot].next=head[s];
57 head[s]=tot++;
58 }
59 void tarjan(int u)//tarjan算法找出图中的所有强连通分支
60 {
61 dfn[u] = low[u]= ++tt;
62 vis[u]=1;
63 s.push(u);
64 int cnt=0;
65 for(int i=head[u];i!=-1;i=edge[i].next)
66 {
67 int v=edge[i].to;
68 if(dfn[v]==-1)
69 {
70 // sum++;
71 tarjan(v);
72 low[u]=min(low[u],low[v]);
73 }
74 else if(vis[v]==1)
75 low[u]=min(low[u],dfn[v]);
76 }
77 if(dfn[u]==low[u])
78 {
79 int x;
80 scc++;
81 do{
82 x=s.top();
83 s.pop();
84 col[x]=scc;
85 vis[x]=0;
86 }while(x!=u);
87 }
88 }
89 bool two_sat(){
90
91 for(int i=0;i<2*n;i++){
92 if(dfn[i]==-1){
93 tarjan(i);
94 }
95 }
96 for(int i=0;i<n;i++){
97 if(col[2*i]==col[2*i+1]){
98 return false;
99 }
100 }
101 return true;
102 }
103 int main()
104 {
105 while(scanf("%d",&n)==1){
106 init();
107 memset(mp,0,sizeof(mp));
108
109 while(!s.empty()){
110 s.pop();
111 }
112 int a,b,c;
113 int x;
114 for(int i=0;i<n;i++){
115 scanf("%d",&x);
116 while(x!=0){
117 x--;
118 mp[i][x]=1;
119 scanf("%d",&x);
120 }
121 }
122 for(int i=0;i<n;i++){
123 for(int j=0;j<n;j++){
124 if(i==j) continue;
125 if(mp[i][j]==0){
126 add(2*i,2*j+1);
127 add(2*j,2*i+1);
128 add(2*i+1,2*j);
129 add(2*j+1,2*i);
130 }
131 }
132 }
133 if(two_sat()){
134 printf("YES\n");
135 }
136 else{
137 printf("NO\n");
138 }
139 }
140 return 0;
141 }