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P2765 魔术球问题

P2765 魔术球问题

题面

题目链接

解法

P2764倒过来。 将和为完全平方数的数对$(a,b)$之间连有向边,大的连向小的,答案就是最长路径覆盖。 枚举答案判断即可。 ### AC代码
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/**
* @file: P2765.cpp
* @author: yaoxi-std
* @url: https://www.luogu.com.cn/problem/P2765
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp &x) {
bool sign = false;
char ch = getchar();
long double tmp = 1;
for (; !isdigit(ch); ch = getchar())
sign |= (ch == '-');
for (x = 0; isdigit(ch); ch = getchar())
x = x * 10 + (ch ^ 48);
if (ch == '.')
for (ch = getchar(); isdigit(ch); ch = getchar())
tmp /= 10.0, x += tmp * (ch ^ 48);
return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
if (x < 0)
putchar('-'), x = -x;
if (x > 9)
write(x / 10);
putchar((x % 10) ^ 48);
}
const int MAXN = 1e4 + 10;
const int MAXM = 2e5 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct Dinic {
struct Edge {
int v, flow;
} edge[MAXM];
int tot = 1, flow = 0;
int head[MAXN], nxt[MAXM], lev[MAXN], cur[MAXN];
void addedge(int u, int v, int flow) {
edge[++tot] = {v, flow};
nxt[tot] = head[u], head[u] = tot;
edge[++tot] = {u, 0};
nxt[tot] = head[v], head[v] = tot;
}
bool bfs(int s, int t) {
memset(lev, -1, sizeof(lev));
queue<int> que;
que.push(s);
lev[s] = 0;
while (!que.empty()) {
int u = que.front();
que.pop();
for (int i = head[u]; i; i = nxt[i]) {
int v = edge[i].v;
if (edge[i].flow && lev[v] == -1) {
lev[v] = lev[u] + 1;
que.push(v);
}
}
}
return lev[t] != -1;
}
int augment(int u, int t, int mx) {
if (u == t || mx == 0)
return mx;
int ret = 0;
for (int &i = cur[u]; i; i = nxt[i]) {
int v = edge[i].v;
if (lev[v] != lev[u] + 1)
continue;
int tmp = augment(v, t, min(mx, edge[i].flow));
mx -= tmp, ret += tmp;
edge[i].flow -= tmp, edge[i ^ 1].flow += tmp;
if (mx == 0)
break;
}
return ret;
}
int maxflow(int s, int t) {
while (bfs(s, t)) {
memcpy(cur, head, sizeof(cur));
flow += augment(s, t, INF);
}
return flow;
}
};
int n, s, t, num, cnt, pre[MAXN], col[MAXN], pt[MAXN][2];
Dinic maxflow;
vector<int> answ[MAXN];
bool issqrt(int x) {
int t = sqrt(x);
return t * t == x;
}
signed main() {
read(n);
s = ++num, t = ++num;
for (int w = 1; ; ++w) {
cnt = 0;
fill(col, col + w, 0);
fill(pre, pre + w, 0);
for (int u = 1; u < w; ++u) {
for (int i = maxflow.head[pt[u][0]]; i; i = maxflow.nxt[i])
if (maxflow.edge[i].flow == INF - 1)
pre[(maxflow.edge[i].v - 1) >> 1] = u;
}
pt[w][0] = ++num, pt[w][1] = ++num;
maxflow.addedge(s, pt[w][0], 1);
maxflow.addedge(pt[w][1], t, 1);
for (int i = 1; i < w; ++i)
if (issqrt(i + w))
maxflow.addedge(pt[w][0], pt[i][1], INF);
int tmp = w - maxflow.maxflow(s, t);
if (tmp > n) {
write(w - 1), putchar('\n');
for (int i = w - 1; i; --i) {
if (!pre[i])
col[i] = ++cnt;
else
col[i] = col[pre[i]];
answ[col[i]].push_back(i);
}
for (int i = 1; i <= cnt; ++i) {
reverse(answ[i].begin(), answ[i].end());
for (int j = 0; j < answ[i].size(); ++j)
write(answ[i][j]), putchar(" \n"[j == answ[i].size() - 1]);
}
break;
}
}
return 0;
}