Hi Guys , I Just Tried to solve the Tanya is 5! problem , hope you might like it , please share if you’ve any better code .
#include <bits/stdc++.h>
#define time privet
#define F first
#define S second
using namespace std;
typedef long double ld;
const int max_n = 42, max_m = 11, max_q = 9, mod = 1000000007, max_log = 20, inf = 1011111111;
const ld eps = 1e-7;
int n, m, b, x[max_n][max_m * 2], cost[max_m], sum[2][max_n + max_m * 2], T, true_m, used1[max_m], used[max_n + max_m * 2], time, ptr[max_m * 2];
vector<pair<pair<int, int>, pair<int, int> > > ans;
pair<int, int> p[max_n + max_m * 2];
vector<pair<int, int> > v[max_n + max_m * 2];
void get_T() {
T = 0;
for (int i = 0; i < n; ++i) {
int sm = 0;
for (int q = 0; q < m; ++q) {
sm += x[i][q];
}
T = max(T, sm);
}
for (int i = 0; i < m; ++i) {
int sm = 0;
for (int q = 0; q < n; ++q) {
sm += x[q][i];
}
T = max(T, sm);
}
}
vector<int> is_exist() {
vector<int> ans;
int res = 0;
for (int i = 0; i < n; ++i) {
int sm = 0;
for (int q = 0; q < m; ++q) {
sm += x[i][q];
}
if (sm == T) return ans;
}
for (int q = 0; q < true_m; ++q) if (!used1[q]){
int sm = 0;
for (int i = 0; i < n; ++i) {
sm += x[i][q];
}
if (sm == T) {
res += cost[q];
ans.push_back(q);
}
}
if (res <= b) {
b -= res;
for (int i = 0; i < ans.size(); ++i) {
used1[ans[i]] = 1;
}
return ans;
}
ans.clear();
return ans;
}
void get_automates() {
get_T();
vector<int> adds = is_exist();
while (adds.size()) {
for (int i = 0; i < adds.size(); ++i) {
int razn = sum[1][adds[i]];
for (int q = v[adds[i]].size() – 1; razn >= v[adds[i]][q].S * 2; –q) {
razn -= v[adds[i]].back().S * 2;
v[m].push_back(v[adds[i]][q]);
sum[1][adds[i]] -= v[m].back().S;
sum[1][m] += v[m].back().S;
swap(x[v[m].back().F][m], x[v[m].back().F][adds[i]]);
v[adds[i]].pop_back();
}
if (razn > 1) {
razn /= 2;
v[m].push_back(make_pair(v[adds[i]].back().F, razn));
v[adds[i]].back().S -= razn;
sum[1][adds[i]] -= razn;
sum[1][m] += razn;
x[v[m].back().F][m] += razn;
x[v[m].back().F][adds[i]] -= razn;
}
ptr[m] = adds[i];
m++;
}
get_T();
adds = is_exist();
}
}
void add_graphs() {
for (int i = 0; i < n; ++i) {
if (sum[0][i] != T) {
v[m + i].push_back(make_pair(i, T – sum[0][i]));
sum[1][m + i] = T – sum[0][i];
sum[0][i] = T;
}
}
for (int i = 0; i < m; ++i) {
if (sum[1][i] != T) {
v[i].push_back(make_pair(n + i, T – sum[1][i]));
sum[0][n + i] = T – sum[1][i];
sum[1][i] = T;
}
}
int i, q;
i = q = n + m – 1;
while (i > n – 1 || q > m – 1) {
if (sum[0][i] == T) {i–; continue;}
if (sum[1][q] == T) {q–; continue;}
int razn = T – max(sum[0][i], sum[1][q]);
//cout << i << ” ” << q << ” ” << razn << “\n”;
sum[0][i] += razn;
sum[1][q] += razn;
v[q].push_back(make_pair(i, razn));
}
}
bool dfs(int x) {
if (used[x]) return false;
used[x] = 1;
for (int i = 0; i < v[x].size(); ++i) {
int to = v[x][i].F;
if (p[to].F == -1 || dfs(p[to].F)) {
p[to] = make_pair(x, i);
return true;
}
}
return false;
}
void get_pairs() {
for (int i = 0; i < n + m; ++i) {
p[i] = make_pair(-1, -1);
}
for (int i = 0; i < n + m; ++i) {
memset(used, 0, sizeof(used));
dfs(i);
}
}
bool is_good() {
for (int i = 0; i < n + m; ++i) if (v[i].size()) return true;
return false;
}
void get_ans() {
while (is_good()) {
get_pairs();
int mn = inf;
for (int i = 0; i < n + m; ++i) {
mn = min(mn, v[p[i].F][p[i].S].S);
}
for (int i = 0; i < n + m; ++i) {
if (i < n && p[i].F < m) {
ans.push_back(make_pair(make_pair(i, ptr[p[i].F]), make_pair(time, mn)));
}
v[p[i].F][p[i].S].S -= mn;
if (v[p[i].F][p[i].S].S == 0) {
v[p[i].F].erase(v[p[i].F].begin() + p[i].S);
}
}
time += mn;
}
}
int main() {
//freopen(“input.txt”, “r”, stdin);
//freopen(“output.txt”, “w”, stdout);
//freopen(“basis.in”, “r”, stdin);
//freopen(“basis.out”, “w”, stdout);
cin >> n >> m >> b;
true_m = m;
for (int i = 0; i < m; ++i) {
scanf(“%d”, &cost[i]);
ptr[i] = i;
}
for (int i = 0; i < n; ++i) {
int k, y, g;
scanf(“%d”, &k);
for (int q = 0; q < k; ++q) {
scanf(“%d%d”, &y, &g);
y–;
x[i][y] = g;
v[y].push_back(make_pair(i, g));
sum[0][i] += g;
sum[1][y] += g;
}
}
get_automates();
add_graphs();
get_ans();
cout << time << “\n”;
for (int i = 0; i < true_m; ++i) {
if (used1[i]) {
cout << “1”;
} else {
cout << “0”;
}
}
cout << “\n” << ans.size() << “\n”;
for (int i = 0; i < ans.size(); ++i) {
cout << ans[i].F.F + 1 << ” ” << ans[i].F.S + 1 << ” ” << ans[i].S.F << ” ” << ans[i].S.S << “\n”;
}
}
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, problem: (E) Tanya is 5! Solution in C/C++. Hi Guys , I Just Tried){kind=link}