2-SAT问题是这样的：有$n$个布尔变量$x_i$，另有$m$个需要满足的条件，每个条件的形式都是“$x_i$为真/假或者$x_j$为真/假”。比如："$x_1$为真或者$x_3$为假“。注意这里的”或“是指两个条件至少有一个是正确的，比如$x_1$和$x_3$一共有$3$中组合满足"$x_1$为真或者$x_3$为假“。2-SAT问题的目标是给每个变量赋值，使得所有条件得到满足。求解2-SAT问题一般比较常见方法是构造一张有向图$G$，其中每个变量$x_i$拆成两个结点$2i$和$2i+1$，分别表示$x_i$为假和$x_i$为真。最后要为每个变量选择其中一个结点标记。比如，若标记了节点$2i$，表示$x_i$为假；如果标记了$2i+1$，表示$x_i$为真。对于“$x_i$为假或者$x_j$为假”这样的条件，我们连一条有向边$2i+1 ightarrow 2j$，表示如果标记节点$2i+1$那么也必须标记结点$j$，同理还需要连一条有向边$2j+1 ightarrow 2i$。对于其他情况，也可以类似连边。换句话说，每个条件对应两条“对称”的边。接下来逐一考虑每个没有赋值的变量，设为$x_i$。我们首先假定它为假，然后标记借点$2_i$，并且沿着有向边标记所有能标记的结点。如果标记过程中发现某个变量对应的两个结点都被标记，则“$x_i$为假”这个假定不成立，需要改成“$x_i$为真”，然后重新标记。注意，该算法无回溯过程。如果当前考虑的变量不管赋值为真还是假都会引起矛盾，可以证明整个2-SAT问题无解（即使调整以前赋值的变量也没用）。这是很显然的，每个变量只会影响到关系到该变量的表达式的取值，因此对于未赋值的变量一定与之前的赋值无关，可以分开考虑，整个问题有解需要满足每个块都有解。下面给出求解2-SAT问题的代码：
1 struct _2_sat{ 2 int n; 3 vector<int> G[maxn << 1]; 4 bool mark[maxn << 1]; 5 int S[maxn << 1], c; 6 bool dfs(int x){ 7 if(mark[x ^ 1]) return 0; 8 if(mark[x]) return 1; 9 mark[x] = 1; 10 S[c++] = x; 11 FOR(i, 0, G[x].size() - 1) if(!dfs(G[x][i])) return 0; 12 return 1; 13 } 14 void init(int n){ 15 this->n = n; 16 FOR(i, 0, 2 * n - 1) G[i].clear(); 17 clr(mark, 0); 18 } 19 //x = xval or y = yval 20 void add_caluse(int x, int xval, int y, int yval){ 21 x = x * 2 + xval, y = y * 2 + yval; 22 G[x ^ 1].pb(y), G[y ^ 1].pb(x); 23 } 24 bool solve(){ 25 for(int i = 0; i < 2 * n; i += 2) if(!mark[i] && !mark[i + 1]){ 26 c = 0; 27 if(!dfs(i)){ 28 while(c > 0) mark[S[--c]] = 0; 29 if(!dfs(i + 1)) return 0; 30 } 31 } 32 return 1; 33 } 34 };
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LA 3713 Astronauts
1 #include <algorithm> 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <queue> 6 #include <map> 7 #include <set> 8 #include <stack> 9 #include <ctime> 10 #include <cmath> 11 #include <iostream> 12 #include <assert.h> 13 #pragma comment(linker, "/STACK:102400000,102400000") 14 #define max(a, b) ((a) > (b) ? (a) : (b)) 15 #define min(a, b) ((a) < (b) ? (a) : (b)) 16 #define mp std :: make_pair 17 #define st first 18 #define nd second 19 #define keyn (root->ch[1]->ch[0]) 20 #define lson (u << 1) 21 #define rson (u << 1 | 1) 22 #define pii std :: pair<int, int> 23 #define pll pair<ll, ll> 24 #define pb push_back 25 #define type(x) __typeof(x.begin()) 26 #define foreach(i, j) for(type(j)i = j.begin(); i != j.end(); i++) 27 #define FOR(i, s, t) for(int i = (s); i <= (t); i++) 28 #define ROF(i, t, s) for(int i = (t); i >= (s); i--) 29 #define dbg(x) std::cout << x << std::endl 30 #define dbg2(x, y) std::cout << x << " " << y << std::endl 31 #define clr(x, i) memset(x, (i), sizeof(x)) 32 #define maximize(x, y) x = max((x), (y)) 33 #define minimize(x, y) x = min((x), (y)) 34 using namespace std; 35 typedef long long ll; 36 const int int_inf = 0x3f3f3f3f; 37 const ll ll_inf = 0x3f3f3f3f3f3f3f3f; 38 const int INT_INF = (int)((1ll << 31) - 1); 39 const double double_inf = 1e30; 40 const double eps = 1e-14; 41 typedef unsigned long long ul; 42 typedef unsigned int ui; 43 inline int readint(){ 44 int x; 45 scanf("%d", &x); 46 return x; 47 } 48 inline int readstr(char *s){ 49 scanf("%s", s); 50 return strlen(s); 51 } 52 53 class cmpt{ 54 public: 55 bool operator () (const int &x, const int &y) const{ 56 return x > y; 57 } 58 }; 59 60 int Rand(int x, int o){ 61 //if o set, return [1, x], else return [0, x - 1] 62 if(!x) return 0; 63 int tem = (int)((double)rand() / RAND_MAX * x) % x; 64 return o ? tem + 1 : tem; 65 } 66 ll ll_rand(ll x, int o){ 67 if(!x) return 0; 68 ll tem = (ll)((double)rand() / RAND_MAX * x) % x; 69 return o ? tem + 1 : tem; 70 } 71 72 void data_gen(){ 73 srand(time(0)); 74 freopen("in.txt", "w", stdout); 75 int kases = 10; 76 printf("%d ", kases); 77 while(kases--){ 78 ll sz = 100000; 79 printf("%d ", sz); 80 FOR(i, 1, sz){ 81 int o = Rand(2, 0); 82 int x = Rand(1e9, 1); 83 int y1 = Rand(1e9, 1), y2 = Rand(1e9, 1); 84 if(o == 0) printf("%d %d %d %d ", x, y1, x, y1); 85 else printf("%d %d %d %d ", y1, x, y2, x); 86 } 87 } 88 } 89 const int maxn = 1e5 + 10; 90 int n, m; 91 ll sum; 92 int id[maxn]; 93 int age[maxn]; 94 int head[maxn << 1]; 95 struct E{ 96 int to, nex; 97 }e[maxn << 2]; 98 bool vis[maxn << 1]; 99 int N; 100 void addE(int x, int y){ 101 e[N].nex = head[x]; 102 e[N].to = y; 103 head[x] = N++; 104 } 105 int stk[maxn], k; 106 bool dfs(int u){ 107 if(vis[u]) return 1; 108 vis[u] = 1; 109 stk[k++] = u; 110 if(vis[u] && vis[u ^ 1]) return 0; 111 for(int i = head[u]; ~i; i = e[i].nex){ 112 int v = e[i].to; 113 if(!dfs(v)) return 0; 114 } 115 return 1; 116 } 117 118 bool solve(int u){ 119 k = 0; 120 if(dfs(2 * u)) return 1; 121 while(k) vis[stk[--k]] = 0; 122 if(dfs(2 * u + 1)) return 1; 123 return 0; 124 } 125 126 int main(){ 127 //data_gen(); return 0; 128 //C(); return 0; 129 int debug = 0; 130 if(debug) freopen("in.txt", "r", stdin); 131 //freopen("out.txt", "w", stdout); 132 while(~scanf("%d%d", &n, &m) && n){ 133 sum = 0; 134 FOR(i, 0, n - 1) age[i] = readint(), sum += age[i]; 135 FOR(i, 0, n - 1) id[i] = (ll)age[i] * n >= sum ? 0 : 1; 136 clr(head, -1), N = 0; 137 FOR(i, 0, m - 1){ 138 int x = readint() - 1, y = readint() - 1; 139 //cc[i][0] = x, cc[i][1] = y; 140 addE(2 * x, 2 * y + 1); 141 if(id[x] == id[y]) addE(2 * x + 1, 2 * y); 142 addE(2 * y, 2 * x + 1); 143 if(id[x] == id[y]) addE(2 * y + 1, 2 * x); 144 } 145 int ok = 1; 146 clr(vis, 0); 147 FOR(i, 0, n - 1) if(!vis[2 * i] && !vis[2 * i + 1] && !solve(i)){ 148 ok = 0; 149 break; 150 } 151 if(!ok) puts("No solution."); 152 else FOR(i, 0, n - 1) putchar(!vis[2 * i + 1] ? 'C' : 'A' + id[i]), putchar(' '); 153 //int res = verdict(); 154 //printf("verdict :: %s", res ? "ok " : "err "); 155 } 156 return 0; 157 }