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QOJ
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 |
---|---|---|---|---|---|---|---|---|---|
#98663 | #6329. Colorful Graph | CSU2023 | WA | 342ms | 19156kb | C++14 | 4.3kb | 2023-04-19 19:33:17 | 2023-04-19 19:33:19 |
Judging History
answer
#include <bits/stdc++.h>
template <class T>
inline void read(T &res)
{
char ch; bool flag = false; res = 0;
while (ch = getchar(), !isdigit(ch) && ch != '-');
ch == '-' ? flag = true : res = ch ^ 48;
while (ch = getchar(), isdigit(ch))
res = res * 10 + ch - 48;
flag ? res = -res : 0;
}
template <class T>
inline void nonnegative_put(T x)
{
if (x > 9)
nonnegative_put(x / 10);
putchar(x % 10 + 48);
}
template <class T>
inline void put(T x)
{
if (x < 0)
x = -x, putchar('-');
nonnegative_put(x);
}
template <class T>
inline void CkMin(T &x, T y) {x > y ? x = y : 0;}
template <class T>
inline void CkMax(T &x, T y) {x < y ? x = y : 0;}
template <class T>
inline T Min(T x, T y) {return x < y ? x : y;}
template <class T>
inline T Max(T x, T y) {return x > y ? x : y;}
template <class T>
inline T Abs(T x) {return x < 0 ? -x : x;}
template <class T>
inline T Sqr(T x) {return x * x;}
//call Sqr((ll)x) when the type of returned value is "long long".
using std::map;
using std::set;
using std::pair;
using std::bitset;
using std::string;
using std::vector;
using std::complex;
using std::multiset;
using std::priority_queue;
typedef long long ll;
typedef long double ld;
typedef complex<ld> com;
typedef pair<int, int> pir;
const ld pi = acos(-1.0);
const ld eps = 1e-8;
const int Maxn = 1e9;
const int Minn = -1e9;
const int mod = 998244353;
const int N = 14e3 + 5;
vector<int> e[N], re[N];
int ans[N], ind[N], dfn[N], low[N], stk[N], col[N];
bool ins[N];
bitset<N> vis[N];
int T_data, n, m, C, top, tis;
const int M = 7e4 + 5;
int nxt[M], to[M], cap[M], adj[N], que[N], cur[N], lev[N];
int src, des, qr, T = 1;
inline void linkArc(int x, int y, int w)
{
nxt[++T] = adj[x]; adj[x] = T; to[T] = y; cap[T] = w;
nxt[++T] = adj[y]; adj[y] = T; to[T] = x; cap[T] = 0;
}
inline bool Bfs()
{
for (int x = 1; x <= des; ++x)
cur[x] = adj[x], lev[x] = -1;
// 初始化具体的范围视建图而定,这里点的范围为 [1,n]
que[qr = 1] = src;
lev[src] = 0;
for (int i = 1; i <= qr; ++i)
{
int x = que[i], y;
for (int e = adj[x]; e; e = nxt[e])
if (cap[e] > 0 && lev[y = to[e]] == -1)
{
lev[y] = lev[x] + 1;
que[++qr] = y;
if (y == des)
return true;
}
}
return false;
}
inline int Dinic(int x, int flow)
{
if (x == des)
return flow;
int y, delta, res = 0;
for (int &e = cur[x]; e; e = nxt[e])
if (cap[e] > 0 && lev[y = to[e]] > lev[x])
{
delta = Dinic(y, Min(flow - res, cap[e]));
if (delta)
{
cap[e] -= delta;
cap[e ^ 1] += delta;
res += delta;
if (res == flow)
break ;
//此时 break 保证下次 cur[x] 仍有机会增广
}
}
if (res != flow)
lev[x] = -1;
return res;
}
inline int maxFlow()
{
int res = 0;
while (Bfs())
res += Dinic(src, Maxn);
return res;
}
inline void Tarjan(int x)
{
dfn[x] = low[x] = ++tis;
stk[++top] = x;
ins[x] = true;
for (int y : e[x])
if (!dfn[y])
{
Tarjan(y);
CkMin(low[x], low[y]);
}
else if (ins[y])
CkMin(low[x], dfn[y]);
if (dfn[x] == low[x])
{
int y;
++C;
do
{
y = stk[top--];
ins[y] = false;
col[y] = C;
}while (y != x);
}
}
inline void dfs(int x)
{
ans[x] = tis;
for (int y : re[x])
if (ind[y])
{
--ind[y];
dfs(y);
return ;
}
}
int main()
{
read(n); read(m);
for (int i = 1, x, y; i <= m; ++i)
{
read(x); read(y);
e[x].emplace_back(y);
}
for (int i = 1; i <= n; ++i)
if (!dfn[i])
Tarjan(i);
for (int x = 1; x <= n; ++x)
{
for (int y : e[x])
if (col[x] != col[y])
vis[col[x]][col[y]] = 1;
e[x].clear();
}
src = (C << 1) | 1, des = src + 1;
for (int x = 1; x <= C; ++x)
{
linkArc(src, x, 1);
linkArc(x + C, des, 1);
linkArc(x + C, x, Maxn);
for (int y = 1; y <= C; ++y)
if (vis[x][y])
linkArc(x, y + C, Maxn);
}
int _ans = maxFlow();
for (int x = 1; x <= C; ++x)
for (int e = adj[x]; e; e = nxt[e])
if (!(e & 1) && to[e] > C && to[e] < src && to[e] - C != x && cap[e] < Maxn)
{
re[x].emplace_back(to[e] - C);
++ind[to[e] - C];
}
tis = 0;
for (int x = 1; x <= C; ++x)
if (!ind[x])
{
++tis;
dfs(x);
}
for (int i = 1; i <= n; ++i)
put(ans[col[i]]), putchar(' ');
putchar('\n');
}
详细
Test #1:
score: 100
Accepted
time: 2ms
memory: 5612kb
input:
5 5 1 4 2 3 1 3 2 5 5 1
output:
2 2 2 1 2
result:
ok AC
Test #2:
score: 0
Accepted
time: 2ms
memory: 5568kb
input:
5 7 1 2 2 1 4 3 5 1 5 4 4 1 4 5
output:
1 1 2 2 2
result:
ok AC
Test #3:
score: 0
Accepted
time: 0ms
memory: 8164kb
input:
8 6 6 1 3 4 3 6 2 3 4 1 6 4
output:
1 1 1 1 2 1 3 4
result:
ok AC
Test #4:
score: -100
Wrong Answer
time: 342ms
memory: 19156kb
input:
7000 6999 4365 4296 2980 3141 6820 4995 4781 24 2416 5844 2940 2675 3293 2163 3853 5356 262 6706 1985 1497 5241 3803 353 1624 5838 4708 5452 3019 2029 6161 3849 4219 1095 1453 4268 4567 1184 1857 2911 3977 1662 2751 6353 6496 2002 6628 1407 4623 425 1331 4445 4277 1259 3165 4994 1044 2756 5788 5496 ...
output:
1 0 0 0 4 2 3 0 0 0 2 2 4 2 0 5 0 6 0 0 0 0 3 0 0 7 8 11 0 9 0 0 10 11 0 0 0 12 3 13 0 0 14 0 15 0 16 3 17 18 0 19 0 4 0 0 2 20 0 21 0 4 4 3 0 22 3 3 0 23 24 0 25 0 26 0 11 0 0 3 3 3 0 0 2 0 0 0 0 3 0 11 246 0 3 0 0 0 27 0 4 5 0 0 0 28 0 4 0 29 4 0 30 0 0 2 0 31 4 3 4 32 33 5 0 0 34 0 0 35 0 0 36 2 ...
result:
wrong answer Integer 0 violates the range [1, 1750]