卡图难题
分析
见《进阶指南》第420
页。
实现
#include <iostream>
#include <cstring>
#include <stack>
using namespace std;
const int N = 2 * 1010, M = 4 * 1000010;
struct edge {
int to, next;
};
edge e[M];
int idx, head[N];
int n, m;
int cnt, dfn[N], low[N];
stack<int> s;
bool in_stack[N];
int tot, scc[N];
void add_edge (int u, int v) {
e[idx].to = v;
e[idx].next = head[u];
head[u] = idx ++;
}
void tarjan (int cur) {
low[cur] = dfn[cur] = ++ cnt;
s.push(cur);
in_stack[cur] = true;
for (int i = head[cur]; i != -1; i = e[i].next) {
int to = e[i].to;
if (dfn[to] == 0) {
tarjan(to);
low[cur] = min(low[cur], low[to]);
} else if (in_stack[to] == true) {
low[cur] = min(low[cur], dfn[to]);
}
}
if (dfn[cur] == low[cur]) {
++ tot;
int t;
do {
t = s.top();
s.pop();
in_stack[t] = false;
scc[t] = tot;
} while (t != cur);
}
}
bool Z_SAT () {
for (int i = 1; i <= 2 * n; ++ i)
if (dfn[i] == 0)
tarjan(i);
for (int i = 1; i <= n; ++ i)
if (scc[i] == scc[i + n])
return false;
return true;
}
int main () {
memset(head, -1, sizeof(head));
scanf("%d%d", &n, &m);
for (int i = 1, a, b, c; i <= m; ++ i) {
char opt[5];
scanf("%d%d%d%s", &a, &b, &c, opt);
if (opt[0] == 'A' && c == 0) {
add_edge(a + n, b);
add_edge(b + n, a);
} else if (opt[0] == 'A' && c == 1) {
add_edge(a, a + n);
add_edge(b, b + n);
} else if (opt[0] == 'O' && c == 0) {
add_edge(a + n, a);
add_edge(b + n, b);
} else if (opt[0] == 'O' && c == 1) {
add_edge(a, b + n);
add_edge(b, a + n);
} else if (c == 0) {
add_edge(a, b);
add_edge(b, a);
add_edge(a + n, b + n);
add_edge(b + n, a + n);
} else {
add_edge(a, b + n);
add_edge(b, a + n);
add_edge(a + n, b);
add_edge(b + n, a);
}
}
printf(Z_SAT() ? "YES" : "NO");
return 0;
}