[Codeforces Round #656 (Div. 3)] (E. Directing Edges)拓撲排序
阿新 • • 發佈:2020-07-24
[Codeforces Round #656 (Div. 3)] (E. Directing Edges)拓撲排序
思路:
首先考慮給定的有向邊構成的圖就存在環的話,輸出NO。
其他情況則一定可以有合法的方案,輸出YES,然後先拿給定的那些有向邊進行拓撲排序,設\(id_i\)為節點\(\mathit i\)在拓撲排序中是第幾個出佇列的(拓撲序的排名)。
然後對於那些無向邊,我們只需要讓拓撲序小的指向拓撲序大的即可。
證明如下:
對於一個無向邊\((u,v)\),
1、如果\((u,v)\)兩節點在原圖(只含有有向邊的那個圖)是聯通的:
即兩者存在至少一個單向路徑,那麼我們讓該邊拓撲序小的節點指向拓撲序大的節點,只是增加了一個和原路徑首位端點相同的單向路徑,不會產生環。
2、如果\((u,v)\)兩節點在原圖中不連通,設它們分別在聯通塊\(A,B\),那麼對於任意\(x\in A,y\in B\),\([id_x>id_y]\)(真值表達式)的值都是一樣的,所以按照拓撲序確定方向也不會產生環。
程式碼:
#include <iostream> #include <cstdio> #include <algorithm> #include <bits/stdc++.h> #define ALL(x) (x).begin(), (x).end() #define sz(a) int(a.size()) #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl #define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c)) #define du2(a,b) scanf("%d %d",&(a),&(b)) #define du1(a) scanf("%d",&(a)); using namespace std; typedef long long ll; ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;} ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;} void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}} void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}} inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;} inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;} void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}} void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}} const int maxn = 1000010; const int inf = 0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ #define DEBUG_Switch 0 int n, m; pii e[maxn]; int op[maxn]; std::vector<int> v[maxn]; int tot; int id[maxn]; int du[maxn]; void init() { repd(i, 1, n) { du[i] = 0; v[i].clear(); } tot = 0; } bool solve() { queue<int> q; int cnt = 0; repd(i, 1, n) { if (du[i] == 0) { q.push(i); cnt++; } } int temp; while (!q.empty()) { temp = q.front(); q.pop(); id[temp] = ++tot; for (auto &y : v[temp]) { if (--du[y] == 0) { q.push(y); cnt++; } } } return cnt == n; } int main() { #if DEBUG_Switch freopen("C:\\code\\input.txt", "r", stdin); #endif //freopen("C:\\code\\output.txt","w",stdout); int t; t = readint(); while (t--) { n = readint(); m = readint(); repd(i, 1, m) { op[i] = readint(); int x = readint(); int y = readint(); e[i] = mp(x, y); if (op[i] == 1) { du[y]++; v[x].push_back(y); } } if (solve()) { printf("YES\n"); repd(i, 1, m) { if (op[i] == 0) { if (id[e[i].fi] > id[e[i].se]) { swap(e[i].fi, e[i].se); } } } repd(i, 1, m) { printf("%d %d\n", e[i].fi, e[i].se ); } } else { printf("NO\n"); } init(); } return 0; }