题目大意

题目又丑又长我就不贴了，说一下大意，有n种菜，m个评委，每一个评委又有两种喜好，每种菜有满汉两种做法，只能选一种。判断是否存在一种方案使得所有评委至少喜欢一种菜品。输入包含多组数据。

题解

显然是2-SAT，注意两种不同做法的菜也要连边。

代码

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000;
vector<int> G[maxn];
vector<int> rG[maxn];
vector<int> sc[maxn];
int vis[maxn], v, cnt[maxn];
vector<int> vs;
void add_edge(int from, int to) {
G[from].push_back(to);
rG[to].push_back(from);
}
void dfs(int u) {
vis[u] = true;
for (int i = 0; i < G[u].size(); i++) {
if (!vis[G[u][i]])
dfs(G[u][i]);
}
vs.push_back(u);
}
void rdfs(int v, int k) {
vis[v] = true;
cnt[v] = k;
for (int i = 0; i < rG[v].size(); i++) {
if (!vis[rG[v][i]])
rdfs(rG[v][i], k);
}
vs.push_back(v);
sc[k].push_back(v);
}
void scc() {
memset(vis, 0, sizeof(vis));
vs.clear();
for (int i = 1; i < v; i++) {
if (!vis[i])
dfs(i);
}
memset(vis, 0, sizeof(vis));
int k = 0;
for (int i = vs.size() - 1; i >= 0; i--) {
if (!vis[vs[i]]) {
rdfs(vs[i], k++);
}
}
}
int n, m, T;
int main() {
// freopen("input", "r", stdin);
scanf("%d", &T);
while (T--) {
memset(cnt, 0, sizeof(cnt));
vs.clear();
for (int i = 0; i < maxn - 2; i++)
sc[i].clear();
scanf("%d %d", &n, &m);
v = 4 * n + 1;
for (int i = 1; i < v; i++) {
G[i].clear(), rG[i].clear();
}
char str[2][maxn];
for (int i = 1; i <= m; i++) {
scanf("%s %s", str[0], str[1]);
int x = 0, y = 0;
for (int i = 1; i < strlen(str[0]); i++)
x = x * 10 + str[0][i] - '0';
for (int i = 1; i < strlen(str[1]); i++)
y = y * 10 + str[1][i] - '0';
if (str[0][0] == 'm')
x = n + x;
if (str[1][0] == 'm')
y = n + y;
add_edge(2 * n + y, x), add_edge(2 * n + x, y);
}
for (int i = 1; i <= n; i++) {
// p:i q:i+n not p:2*n+i not q:3*n+i
add_edge(i, 3 * n + i), add_edge(i + n, 2 * n + i);
}
scc();
int flag = 0;
for (int i = 1; i <= n; i++) {
if (cnt[i] == cnt[i + n]) {
printf("BAD\n");
flag = 1;
break;
}
}
if (!flag)
printf("GOOD\n");
}
}