BZOJ2199 [Usaco2011 Jan]奶牛议会

首先激励一个2-SAT的裸模型，然后发现。。。tarjan没法判断‘?‘的情况

于是暴力对每一个议案check一下，直接dfs即可

  1 /**************************************************************
  2     Problem: 2199
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:160 ms
  7     Memory:884 kb
  8 ****************************************************************/
  9
 10 #include <cstdio>
 11 #include <cstring>
 12 #include <algorithm>
 13
 14 using namespace std;
 15 const char ch[3] = {‘?‘, ‘N‘, ‘Y‘};
 16 const int N = 2005;
 17 const int M = N << 2;
 18
 19 struct edge {
 20   int next, to;
 21   edge() {}
 22   edge(int _n, int _t) : next(_n), to(_t) {}
 23 } e[M];
 24
 25 int n, m;
 26 int first[N], tot;
 27 bool vis[N];
 28 int ans[N];
 29
 30 int read() {
 31   int x = 0;
 32   char ch = getchar();
 33   while (ch < ‘0‘ || ‘9‘ < ch)
 34     ch = getchar();
 35   while (‘0‘ <= ch && ch <= ‘9‘)
 36     (x *= 10) += ch - ‘0‘, ch = getchar();
 37   return x;
 38 }
 39
 40 int get() {
 41   int x = read();
 42   char ch = getchar();
 43   while (ch != ‘Y‘ && ch != ‘N‘)
 44     ch = getchar();
 45   if (ch == ‘Y‘) --(x <<= 1);
 46   else x <<= 1;
 47   return x;
 48 }
 49
 50 void add_edge(int x, int y) {
 51   e[++tot] = edge(first[x], y);
 52   first[x] = tot;
 53 }
 54
 55 #define y e[x].to
 56 void dfs(int p) {
 57   int x;
 58   vis[p] = 1;
 59   for (x = first[p]; x; x = e[x].next)
 60     if (!vis[y]) dfs(y);
 61 }
 62 #undef y
 63
 64 bool check(int p) {
 65   int i;
 66   memset(vis, 0, sizeof(vis));
 67   dfs(p);
 68   for (i = 1; i <= n; ++i)
 69     if (vis[2 * i] && vis[2 * i - 1]) return 0;
 70   return 1;
 71 }
 72
 73
 74 int main() {
 75   int i, a, b, c, d, p, q;
 76   n = read(), m = read();
 77   for (i = 1; i <= m; ++i) {
 78     a = get(), c = get();
 79     if (a & 1) b = a + 1;
 80     else b = a - 1;
 81     if (c & 1) d = c + 1;
 82     else d = c - 1;
 83     add_edge(b, c);
 84     add_edge(d, a);
 85   }
 86   for (i = 1; i <= n; ++i) {
 87     p = check(2 * i - 1);
 88     q = check(2 * i);
 89     if (!p && !q) {
 90       puts("IMPOSSIBLE");
 91       return 0;
 92     }
 93     else if (p && q) ans[i] = 0;
 94     else if (!p) ans[i]= 1;
 95     else ans[i] = 2;
 96   }
 97   for (i = 1; i <= n; ++i)
 98     putchar(ch[ans[i]]);
 99   puts("");
100   return 0;
101 }