uva 1086 - The Ministers' Major Mess（2 SAT）

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题目链接：uva 1086 - The Ministers' Major Mess

枚举每个点，判断是否y，n都存在解，如果都存在即为？, 最后做一遍2SAT即可。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>using namespace std;
const int maxn = 105;struct TwoSAT {int n, S[maxn * 2], C;bool mark[maxn * 2], must[maxn * 2];vector<int> G[maxn * 2];void init (int n) {this->n = n;memset(must, 0, sizeof(must));for (int i = 0; i < n*2; i++) G[i].clear();}void clear() { memcpy(mark, must, sizeof(must)); }void addClause(int x, int xflag, int y, int yflag) {x = x * 2 + xflag;y = y * 2 + yflag;G[x^1].push_back(y);G[y^1].push_back(x);}void draw(int u) {must[u] = true;for (int i = 0; i < G[u].size(); i++)if (!must[G[u][i]]) draw(G[u][i]);}bool dfs (int u) {if (mark[u^1]) return false;if (mark[u]) return true;mark[u] = true;S[C++] = u;for (int i = 0; i < G[u].size(); i++)if (!dfs(G[u][i])) return false;return true;}bool solve () {for (int i = 0; i < 2*n; i += 2) {if (mark[i] && mark[i+1]) return false;if (!mark[i] && !mark[i+1]) {C = 0;if (!dfs(i)) {while (C) mark[S[--C]] = false;if (!dfs(i+1)) return false;}}}return true;}
}solver;int N, M, T[maxn];void init () {int k, x[10], y[10];char s[10];solver.init(N);memset(T, 0, sizeof(T));while (M--) {scanf("%d", &k);for (int i = 0; i < k; i++) {scanf("%d%s", &x[i], s);x[i]--; y[i] = s[0] == 'y' ? 0 : 1;}if (k >= 3) {for (int i = 0; i < k; i++)for (int j = i + 1; j < k; j++)solver.addClause(x[i], y[i], x[j], y[j]);} else {for (int i = 0; i < k; i++)solver.must[x[i]*2+y[i]] = true;}}
}int main () {int cas = 1;while (scanf("%d%d", &N, &M) == 2 && N + M) {init();for (int i = 0; i < 2*N; i++)if (solver.must[i]) solver.draw(i);for (int i = 0; i < 2*N; i += 2) {if (solver.must[i] || solver.must[i+1]) continue;solver.clear();solver.mark[i] = true;bool flag1 = solver.solve();solver.clear();solver.mark[i+1] = true;bool flag2 = solver.solve();if (flag1 && flag2) T[i/2] = 1;}solver.clear();printf("Case %d: ", cas++);if (!solver.solve()) printf("impossible\n");else {for (int i = 0; i < N; i++) {if (T[i]) printf("?");else if (solver.mark[i*2]) printf("y");else printf("n");}printf("\n");}}return 0;
}

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