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

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;
}