平面图中E ≤ V*2-6..
一个圈上2个点的边可以是在外或者内, 经典的2sat问题..
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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
using namespace std;
#define U(x) U[r[x]]
#define V(x) V[r[x]]
#define H(x) H[r[x]]
const int maxn = 20009;
struct edge {
int to;
edge* next;
} E[1000000], *pt, *head[maxn];
void AddEdge(int u, int v) {
pt->to = v; pt->next = head[u]; head[u] = pt++;
}
int Low[maxn], Dfn[maxn], Scc[maxn], CK, scc_n;
int U[maxn], V[maxn], H[maxn], P[maxn], _P[maxn], r[maxn], n;
int N, M, T;
stack<int> S;
void Init() {
pt = E;
memset(head, 0, sizeof head);
scanf("%d%d", &N, &M);
for(int i = 0; i < M; i++)
scanf("%d%d", U + i, V + i);
for(int i = 0; i < N; i++) {
scanf("%d", _P + i);
P[_P[i]] = i;
}
for(int i = 0; i < N; i++)
H[_P[i]] = _P[(i + 1) % N];
n = scc_n = CK = 0;
memset(Dfn, 0, sizeof Dfn);
memset(Scc, 0, sizeof Scc);
while(!S.empty()) S.pop();
}
bool chk(int l, int r, int _l, int _r) {
if(l > r) swap(l, r);
if(_l > _r) swap(_l, _r);
return (l < _l && _l < r && r < _r) || (_l < l && l < _r && _r < r);
}
void Tarjan(int x) {
Dfn[x] = Low[x] = ++CK;
S.push(x);
for(edge* e = head[x]; e; e = e->next) if(!Dfn[e->to]) {
Tarjan(e->to);
Low[x] = min(Low[x], Low[e->to]);
} else if(!Scc[e->to])
Low[x] = min(Low[x], Dfn[e->to]);
if(Dfn[x] == Low[x]) {
int t; scc_n++;
do {
t = S.top(); S.pop();
Scc[t] = scc_n;
} while(t != x);
}
}
bool Solve() {
if(M > 3 * N - 6) return false;
for(int i = 0; i < M; i++)
if(V[i] != H[U[i]] && U[i] != H[V[i]]) r[n++] = i;
for(int i = 0; i < n; i++)
for(int j = 0; j < i; j++)
if(chk(P[U(i)], P[V(i)], P[U(j)], P[V(j)])) {
AddEdge(i * 2, j * 2 + 1);
AddEdge(i * 2 + 1, j * 2);
AddEdge(j * 2, i * 2 + 1);
AddEdge(j * 2 + 1, i * 2);
}
for(int i = 0; i < 2 * n; i++) {
if(!Dfn[i]) Tarjan(i);
}
for(int i = 0; i < n; i++)
if(Scc[i * 2] == Scc[i * 2 + 1]) return false;
return true;
}
int main() {
scanf("%d", &T);
while(T--) {
Init();
puts(Solve() ? "YES" : "NO");
}
return 0;
}
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1997: [Hnoi2010]Planar
Time Limit: 10 Sec Memory Limit: 64 MBSubmit: 1183 Solved: 458
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