The Bandulu Space Agency (BSA) has plans for the following three space
missions:
- Mission A: Landing on Ganymede, the largest moon of Jupiter.
- Mission B: Landing on Callisto, the second largest moon of Jupiter.
- Mission C: Landing on Titan, the largest moon of Saturn.
Your task is to assign a crew for each mission. BSA has trained a number of
excellent astronauts; everyone of them can be assigned to any mission. However,
if two astronauts hate each other, then it is not wise to put them on the same
mission. Furthermore, Mission A is clearly more prestigious than Mission B; who
would like to go to the second largest moon if there is also a mission to the
largest one? Therefore, the assignments have to be done in such a way that only
young, inexperienced astronauts go to Mission B, and only senior astronauts are
assigned to Mission A. An astronaut is considered young if their age is less than the average age of
the astronauts and an astronaut is senior if their age is at least the averageage. Every
astronaut can be assigned to Mission C, regardless of their age (but you must
not assign two astronauts to the same mission if they hate each other).
Input
The input contains several blocks of test cases. Each case begins with a line
containing two integers 1
n100000 <tex2html_verbatim_mark>and 1m100000 <tex2html_verbatim_mark>. The
number n <tex2html_verbatim_mark>is the number
of astronauts. The next n <tex2html_verbatim_mark>lines specify
the age of the n<tex2html_verbatim_mark>astronauts; each
line contains a single integer number between 0 and 200. The next m <tex2html_verbatim_mark>lines contains
two integers each, separated by a space. A line containing i <tex2html_verbatim_mark>and j <tex2html_verbatim_mark>(1i, jn) <tex2html_verbatim_mark>means that
the i <tex2html_verbatim_mark>-th astronaut
and the j <tex2html_verbatim_mark>-th astronaut
hate each other.
The input is terminated by a block with n = m =
0 <tex2html_verbatim_mark>.
Output
For each test case, you have to output n lines, each containing a single
letter. This letter is either `A‘, `B‘, or `C‘.
The i <tex2html_verbatim_mark>-th line
describes which mission the i <tex2html_verbatim_mark>-th astronaut
is assigned to. Astronauts that hate each other should not be assigned to the
same mission, only young astronauts should be assigned to Mission B and only
senior astronauts should be assigned to Mission A. If there is no such
assignment, then output the single line `No solution.‘ (without
quotes).
Sample
Input
16 20
21
22
23
24
25
26
27
28
101
102
103
104
105
106
107
108
1 2
3 4
5 6
7 8
9 10
11 12
13 14
15 16
1 10
2 9
3 12
4 11
5 14
6 13
7 16
8 15
1 12
1 13
3 16
6 15
0 0
Sample
Output
B
C
C
B
C
B
C
B
A
C
C
A
C
A
C
A
拆成两个分组走2-sat.
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <stack>
6 #include <vector>
7
8 using namespace std;
9
10 const int MAX_N = 1e5 + 5;
11 int N,M;
12 int low[MAX_N * 2],pre[MAX_N * 2],cmp[MAX_N * 2];
13 int age[MAX_N];
14 vector <int> G[2 * MAX_N];
15 stack <int> S;
16 int dfs_clock,scc_cnt;
17
18 void dfs(int u) {
19 pre[u] = low[u] = ++dfs_clock;
20 S.push(u);
21 for(int i = 0; i < G[u].size(); ++i) {
22 int v = G[u][i];
23 if(!pre[v]) {
24 dfs(v);
25 low[u] = min(low[u],low[v]);
26 } else if(!cmp[v]) {
27 low[u] = min(low[u],pre[v]);
28 }
29 }
30
31 if(pre[u] == low[u]) {
32 ++scc_cnt;
33 for(;;) {
34 int x = S.top(); S.pop();
35 cmp[x] = scc_cnt;
36 if(x == u) break;
37 }
38 }
39 }
40
41 bool scc() {
42 dfs_clock = scc_cnt = 0;
43 memset(cmp,0,sizeof(cmp));
44 memset(pre,0,sizeof(pre));
45
46 for(int i = 1; i <= 2 * N; ++i) {
47 if(!pre[i]) dfs(i);
48 }
49
50 for(int i = 1; i <= N; ++i) {
51 if(cmp[i] == cmp[N + i]) return false;
52 }
53
54 return true;
55 }
56 int main()
57 {
58 freopen("sw.in","r",stdin);
59 while(~scanf("%d%d",&N,&M) && (N || M)) {
60 int sum = 0;
61 for(int i = 1; i <= N; ++i) {
62 scanf("%d",&age[i]);
63 sum += age[i];
64 }
65 for(int i = 1; i <= 2 * N; ++i) G[i].clear();
66
67 for(int i = 1; i <= M; ++i) {
68 int u,v;
69 scanf("%d%d",&u,&v);
70 if(age[u] * N >= sum && age[v] * N < sum
71 || age[u] * N < sum && age[v] * N >= sum) {
72 G[v + N].push_back(u);
73 G[u + N].push_back(v);
74 } else {
75
76 G[u].push_back(v + N);
77 G[v].push_back(u + N);
78 G[v + N].push_back(u);
79 G[u + N].push_back(v);
80 }
81
82 }
83
84 if(!scc()) {
85 printf("No solution.\n");
86 } else {
87 for(int i = 1; i <= N; ++i) {
88 if(age[i] * N >= sum) {
89 printf("%c\n",cmp[i] < cmp[i + N] ? ‘A‘ : ‘C‘);
90 } else {
91 printf("%c\n",cmp[i] < cmp[i + N] ? ‘B‘ : ‘C‘);
92 }
93 }
94 }
95 }
96
97 return 0;
98 }