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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#695637 | #7107. Chaleur | Mine_King# | WA | 180ms | 11400kb | C++14 | 2.4kb | 2024-10-31 20:29:42 | 2024-10-31 20:29:44 |
Judging History
answer
// 長い夜の終わりを信じながら
// Think twice, code once.
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define eputchar(c) putc(c, stderr)
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#define eputs(str) fputs(str, stderr), putc('\n', stderr)
using namespace std;
int T, n, m, vis[100005];
struct graph {
int tot, hd[100005];
int nxt[200005], to[200005];
void clear(int n) {tot = 0; memset(hd, 0, sizeof(int) * n); return;}
void add(int u, int v) {
nxt[++tot] = hd[u];
hd[u] = tot;
to[tot] = v;
return;
}
} g;
int l[100005], p[100005], pos[100005], q[100005];
vector<int> vec[100005];
int deg[100005], cnt[100005];
int main() {
scanf("%d", &T);
while (T--) {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) vis[i] = deg[i] = cnt[i] = 0;
g.clear(n + 5);
for (int i = 1; i <= m; i++) {
int u, v;
scanf("%d%d", &u, &v);
g.add(u, v), g.add(v, u);
deg[u]++, deg[v]++;
}
for (int i = 1; i <= n; i++) l[i] = 0, pos[i] = vec[0].size(), vec[0].push_back(i);
for (int i = 1, j = 0; i <= n; i++, j++) {
while (vec[j].empty()) j--;
int u = p[i] = vec[j].back();
vec[j].pop_back();
pos[u] = -1;
for (int k = g.hd[u]; k; k = g.nxt[k])
if (pos[g.to[k]] != -1) {
int v = g.to[k];
pos[vec[l[v]].back()] = pos[v];
swap(vec[l[v]][pos[v]], vec[l[v]].back());
vec[l[v]].pop_back();
pos[v] = vec[++l[v]].size();
vec[l[v]].push_back(v);
}
}
reverse(p + 1, p + n + 1);
for (int i = 1; i <= n; i++) q[p[i]] = i;
int mx = 0, id = 0;
for (int i = 1; i <= n; i++) {
int num = 1;
for (int j = g.hd[p[i]]; j; j = g.nxt[j]) num += q[g.to[j]] > i;
if (num > mx) mx = num, id = p[i];
}
int ans1 = 1, ans2 = 1;
vis[id] = 1;
for (int i = g.hd[id]; i; i = g.nxt[i])
if (q[g.to[i]] > q[id]) vis[g.to[i]] = 1;
for (int i = 1; i <= n; i++)
if (!vis[i]) {
ans1 += deg[i] == mx - 1;
for (int j = g.hd[i]; j; j = g.nxt[j])
if (vis[g.to[j]]) cnt[g.to[j]]++;
}
for (int i = 1; i <= n; i++) ans2 += cnt[i] == 1;
for (int i = 1; i <= n; i++)
if (vis[i] && !cnt[i]) {
ans2 = 0;
for (int i = 1; i <= n; i++) ans2 += vis[i] && !cnt[i];
break;
}
printf("%d %d\n", ans1, ans2);
}
return 0;
}
/*
3
3 2
1 2
2 3
6 6
1 2
2 3
1 3
1 4
2 5
3 6
4 1
1 2
*/
详细
Test #1:
score: 100
Accepted
time: 2ms
memory: 10044kb
input:
3 3 2 1 2 2 3 6 6 1 2 2 3 1 3 1 4 2 5 3 6 4 1 1 2
output:
2 1 1 4 1 2
result:
ok 3 lines
Test #2:
score: -100
Wrong Answer
time: 180ms
memory: 11400kb
input:
2231 1 0 5 7 4 1 3 4 3 1 3 5 4 2 3 2 4 5 5 4 2 1 2 5 2 4 2 3 5 10 3 2 2 5 1 4 4 2 4 5 1 2 1 3 3 5 3 4 1 5 5 10 1 3 2 4 1 4 5 2 2 3 1 5 5 4 1 2 3 4 5 3 5 9 2 5 3 5 2 3 2 1 4 3 3 1 4 1 4 5 2 4 5 4 4 2 4 1 4 5 4 3 5 9 4 1 4 5 3 4 2 4 2 1 3 1 2 5 3 5 3 2 5 4 2 5 2 3 2 1 2 4 5 9 5 2 1 3 4 3 1 2 5 4 4 2 5...
output:
1 1 3 1 4 1 1 5 1 5 2 1 4 1 2 1 4 1 2 1 2 1 3 1 4 1 4 1 1 5 2 1 4 1 1 5 1 5 1 5 3 1 4 1 4 1 4 1 3 1 3 1 4 1 4 1 2 1 4 1 4 1 1 5 1 5 2 1 4 1 4 1 4 1 3 1 2 1 4 1 2 1 4 1 4 1 4 1 3 1 1 5 4 1 4 1 1 5 2 1 4 1 2 1 2 1 1 5 4 1 1 5 3 1 4 1 1 5 2 1 1 5 3 1 3 1 1 5 3 1 3 1 2 1 1 5 4 1 3 1 1 5 2 1 3 1 2 1 2 1 ...
result:
wrong answer 1021st lines differ - expected: '4 1', found: '3 1'