Codeforces Round #599 (Div. 2) E. Sum Balance
这题写起来真的有点麻烦，按照官方题解的写法
先建图，然后求强连通分量，然后判断掉不符合条件的换
最后做dp转移即可
虽然看起来复杂度很高，但是n只有15，所以问题不大
#include <iostream>
#include <fstream>
#include <vector>
#include <set>
#include <map>
#include <bitset>
#include <algorithm>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <functional>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <deque>
#include <stack>
#include <complex>
#include <cassert>
#include <random>
#include <cstring>
#include <numeric>
#define ll long long
#define ld long double
#define null NULL
#define all(a) a.begin(), a.end()
#define forn(i, n) for (int i = 0; i < n; ++i)
#define sz(a) (int)a.size()
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define bitCount(a) __builtin_popcount(a)
template<class T> int gmax(T &a, T b) { if (b > a) { a = b; return 1; } return 0; }
template<class T> int gmin(T &a, T b) { if (b < a) { a = b; return 1; } return 0; }
using namespace std;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string) s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) { return "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; }
template <typename A>
string to_string(A v) { bool first = true; string res = "{"; for (const auto &x : v) { if (!first) { res += ", "; } first = false; res += to_string(x); } res += "}"; return res; }
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) { cerr << " " << to_string(H); debug_out(T...); }
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif
const int MAXN = 75005;
vector<int> graph[MAXN];
vector<int> group[MAXN];
int dfn[MAXN], low[MAXN];
int dcnt;
int col[MAXN + 5], ccnt;
bool vis[MAXN + 5];
int stk[MAXN + 5], tp;
vector<pair<int, int>> dp[32768];
void Tarjan_scc(int u) {
dfn[u] = ++dcnt, low[u] = dcnt, vis[u] = true;
stk[++tp] = u;
for(int i = 0; i < (int)graph[u].size(); i++) {
int v = graph[u][i];
if(!dfn[v]) {
Tarjan_scc(v);
low[u] = min(low[u], low[v]);
} else if(vis[v]) low[u] = min(low[u], low[v]);
}
if(dfn[u] == low[u]) {
++ccnt;
while(true) {
col[stk[tp]] = ccnt;
vis[stk[tp]] = false;
if(stk[tp--] == u)
break;
}
}
}
int main() {
int k;
while(~scanf("%d", &k)) {
tp = -1; ccnt = 0; dcnt = 0;
for(int i = 0; i < MAXN; ++i) {
dfn[i] = 0;
graph[i].clear();
group[i].clear();
}
vector<ll> sum;
vector<pair<int, int> > vc;
map<ll, pair<int, int> > mp;
int tot = 0;
ll allSum = 0;
for(int i = 0; i < k; ++i) {
int x; scanf("%d", &x);
ll tmpSum = 0;
for(int j = 0; j < x; ++j) {
int y; scanf("%d", &y);
vc.push_back(make_pair(y, i));
mp[y] = make_pair(tot, i);
tot ++;
tmpSum += y;
}
sum.push_back(tmpSum);
allSum += tmpSum;
}
// debug(allSum);
if(allSum % k) {
printf("No\n");
continue;
}
allSum /= k;
set<int> selfCircle;
set<pair<int, int>> hasEdge;
// debug(allSum);
for(int i = 0, len = vc.size(); i < len; ++i) {
ll searchNum = allSum - sum[vc[i].second] + vc[i].first;
if(mp.find(searchNum) == mp.end()) continue;
else if( mp[searchNum].second == vc[i].second && mp[searchNum].first != i) {
// solve specfial condition
continue;
} else if(mp[searchNum].first == i) {
// debug(i);
selfCircle.insert(i);
}
graph[i].push_back(mp[searchNum].first);
hasEdge.insert(make_pair(i, mp[searchNum].first));
debug(i, mp[searchNum].first);
}
for(int i = 0; i < tot; ++i) {
if(!dfn[i]) Tarjan_scc(i);
}
for(int i = 0; i < tot; ++i) {
// printf("%d ", col[i]);
group[col[i]].push_back(i);
}
// printf("\n");
for(int i = 1; i <= ccnt; ++i) {
// debug(i, group[i].size());
if(group[i].size() == 1 && selfCircle.count(group[i][0])) {
int id = group[i][0];
vector<pair<int, int> > tmpPair;
tmpPair.push_back(make_pair(vc[id].first, vc[id].second + 1));
dp[1<<vc[id].second] = tmpPair;
// debug(dp[1<<vc[id].second]);
}
else if(group[i].size() > 1) {
vector<pair<int, int> > tmpPair;
int tmp = 0;
int len = group[i].size();
bool suc = true;
for(int j = 0; j < len; ++j) {
int to = group[i][j];
if(tmp & (1<<vc[to].second)) { suc = false; break; }
tmp |= 1<<vc[to].second;
}
if(suc == false) {
continue;
}
for(int j = 0; j < len; ++j) {
for(int k = 0; k < len; ++k) {
int fr = group[i][j]; int to = group[i][k];
if(hasEdge.count(make_pair(fr, to))) {
tmpPair.push_back(make_pair(vc[to].first, vc[fr].second + 1));
}
}
}
dp[tmp] = tmpPair;
// debug(dp[tmp], tmp);
}
}
for(int i = 0; i < (1<<k); ++i) {
for (int s=(i-1)&i; s; s=(s-1)&i) {
int t = i ^ s;
if(dp[s].size() > 0 && dp[t].size() > 0) {
dp[i] = dp[s];
dp[i].insert(dp[i].end(), dp[t].begin(), dp[t].end());
break;
}
}
}
auto cmp = [&](pair<int, int> &A, pair<int, int> &B) {
return mp[A.first].second < mp[B.first].second;
};
int end = (1<<k) - 1;
if(dp[end].size() > 0 ) {
printf("Yes\n");
// debug(dp[end]);
sort(dp[end].begin(), dp[end].end(), cmp);
for(int i = 0; i < k; ++i) {
printf("%d %d\n", dp[end][i].first, dp[end][i].second);
}
} else printf("No\n");
}
return 0;
}
