結果
| 問題 |
No.3173 じゃんけんの勝ちの回数
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2025-06-06 21:42:44 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 40 ms / 2,000 ms |
| コード長 | 2,800 bytes |
| コンパイル時間 | 1,937 ms |
| コンパイル使用メモリ | 197,460 KB |
| 実行使用メモリ | 18,256 KB |
| 最終ジャッジ日時 | 2025-06-06 21:42:50 |
| 合計ジャッジ時間 | 5,467 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 1 |
| other | AC * 33 |
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:91:50: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
91 | for (int i = 1; i < 4; i++) scanf("%d", a + i), add(0, i, a[i], 0);
| ~~~~~^~~~~~~~~~~~~
main.cpp:92:50: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
92 | for (int i = 1; i < 4; i++) scanf("%d", b + i), add(i + 3, et, b[i], 0);
| ~~~~~^~~~~~~~~~~~~
ソースコード
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
const int N = 2000008, MOD = 998244353, INF = 0x3f3f3f3f;
ll res;
int n, m, cnt, w[N];
int a[4], b[4];
int e[N], ne[N], h[N], idx, d[N], inc[N], pre[N], v[N];
bool st[N];
int s = 0, et = 7;
void add(int a, int b, int c, int d) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, v[idx] = d, h[a] = idx++;
e[idx] = a, ne[idx] = h[b], w[idx] = 0, v[idx] = -d, h[b] = idx++;
}
bool spfa() {
queue<int> q;
// memset(d, -0x3f, sizeof(int) * (n + 10));
memset(d, 0x3f, sizeof(int) * (20));
q.push(s), d[s] = 0, inc[s] = INF;
while (q.size()) {
int u = q.front();
q.pop(); st[u] = 0;
for (int i = h[u]; ~i; i = ne[i]) {
int j = e[i];
if (w[i] && d[j] > d[u] + v[i]) {
d[j] = d[u] + v[i], pre[j] = i, inc[j] = min(inc[u], w[i]);
if (!st[j]) q.push(j), st[j] = 1;
}
}
}
return d[et] != INF;
}
void ek(int& f, int& c) {
f = c = 0;
while (spfa()) {
int t = inc[et];
f += t, c += t * d[et];
for (int i = et; i != s; i = e[pre[i] ^ 1]) {
w[pre[i]] -= t, w[pre[i] ^ 1] += t;
}
}
}
bool spfa2() {
queue<int> q;
memset(d, -0x3f, sizeof(int) * (n + 10));
q.push(s), d[s] = 0, inc[s] = INF;
while (q.size()) {
int u = q.front();
q.pop(); st[u] = 0;
for (int i = h[u]; ~i; i = ne[i]) {
int j = e[i];
if (w[i] && d[j] < d[u] + v[i]) {
d[j] = d[u] + v[i], pre[j] = i, inc[j] = min(inc[u], w[i]);
if (!st[j]) q.push(j), st[j] = 1;
}
}
}
return d[et] >= 0;
}
void ek2(int& f, int& c) {
f = c = 0;
while (spfa2()) {
int t = inc[et];
f += t, c += t * d[et];
for (int i = et; i != s; i = e[pre[i] ^ 1]) {
w[pre[i]] -= t, w[pre[i] ^ 1] += t;
}
}
}
int main() {
int T;
cin >> T;
while (T--) {
memset(h, -1, sizeof(int) * (20));
idx = 0;
for (int i = 1; i < 4; i++) scanf("%d", a + i), add(0, i, a[i], 0);
for (int i = 1; i < 4; i++) scanf("%d", b + i), add(i + 3, et, b[i], 0);
for (int i = 1; i < 4; i++) {
for (int j = 1; j < 4; j++) {
if (i + 1 == j || (i == 3 && j == 1)) add(i, j + 3, INF, 1);
else add(i, j + 3, INF, 0);
}
}
int f, c;
ek(f, c);
printf("%d ", c);
memset(h, -1, sizeof(int) * (20));
idx = 0;
for (int i = 1; i < 4; i++) add(0, i, a[i], 0);
for (int i = 1; i < 4; i++) add(i + 3, et, b[i], 0);
for (int i = 1; i < 4; i++) {
for (int j = 1; j < 4; j++) {
if (i + 1 == j || (i == 3 && j == 1)) add(i, j + 3, INF, 1);
else add(i, j + 3, INF, 0);
}
}
ek2(f, c);
printf("%d\n", c);
}
return 0;
}