Codeforces Round #380 (Div. 1, Rated, Based on Technocup 2017 – Elimination Round 2), problem: (E) Tanya is 5! Solution in C/C++

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”;
}
}

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”;
}
}

More from author

LEAVE A REPLY

Please enter your comment!
Please enter your name here

Related posts

Advertismentspot_img

Latest posts

I Tried Phone Stand For Desk For 30 Days Here Is the Honest Truth

📢 As an Amazon Associate, I earn from qualifying purchases.If you had told me six months ago that a single purchase would change how...

Stop Buying Cheap Alternatives This Wireless Earbuds Accessories Is the Real Deal

📢 As an Amazon Associate, I earn from qualifying purchases.I stumbled across this while desperately trying to fix a recurring problem in my routine....

Where the OKC Thunder Rank Among the Top 10 NBA Teams Entering August

As the offseason progresses into August, the NBA landscape continues to undergo dramatic transformations. Major headline-grabbing moves, including LeBron James making his historic transition...

Want to stay up to date with the latest news?

We would love to hear from you! Please fill in your details and we will stay in touch. It's that simple!