题意
给 \(n\) 个点两两求最小割,问不同的最小割的数量。
Sol
最小割树。
每次最小割完,对于源点集和汇点集分别再做一遍最小割。
这样递归下去对于每次的源点和汇点连边,边权为最小割的值。
Code
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <queue>
#include <vector>
#include <bitset>
#include <tuple>
#include <unordered_map>
#define pii pair <int, int>
using namespace std;
#ifdef ONLINE_JUDGE
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;
#endif
int read() {
int p = 0, flg = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') flg = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
p = p * 10 + c - '0';
c = getchar();
}
return p * flg;
}
void write(int x) {
if (x < 0) {
x = -x;
putchar('-');
}
if (x > 9) {
write(x / 10);
}
putchar(x % 10 + '0');
}
#define fi first
#define se second
const int N = 1005, M = 4e4 + 5, inf = 2e9;
namespace G {
array <int, N> fir;
array <int, M> nex, to, cap;
int cnt = 1;
void add(int x, int y, int z) {
cnt++;
nex[cnt] = fir[x];
to[cnt] = y;
cap[cnt] = z;
fir[x] = cnt;
}
void _add(int x, int y, int z) {
add(x, y, z);
add(y, x, 0);
}
}
namespace Mfl {
array <int, N> dis, cur;
queue <int> q;
bool bfs(pii st) {
dis.fill(-1), dis[st.fi] = 0;
q.push(st.fi);
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = G::fir[u]; i; i = G::nex[u]) {
if (!G::cap[i] || ~dis[G::to[i]]) continue;
dis[G::to[i]] = dis[u] + 1, q.push(G::to[i]);
}
}
return ~dis[st.se];
}
int dfs(int x, int augcd, pii st) {
if (x == st.se) return augcd;
int augc = augcd;
for (int &i = cur[x]; i; i = G::nex[i]) {
if (!G::cap[i] || dis[G::to[i]] <= dis[x]) continue;
int flow = dfs(G::to[i], min(augc, G::cap[i]), st);
augc -= flow;
G::cap[i] -= flow, G::cap[i ^ 1] += flow;
if (!augc) break;
}
return augcd - augc;
}
int dinic(pii st) {
int ans = 0;
while (bfs(st)) {
// copy(G::fir.begin(), G::fir.end(), cur.begin());
cur = G::fir;
ans += dfs(st.fi, inf, st);
}
return ans;
}
}
array <tuple <int, int, int>, M> edge;
namespace Mft {
bitset <N> vis;
void dfs(int x) {
vis[x] = 1;
for (int i = G::fir[x]; i; i = G::nex[i]) {
if (!G::cap[i] || vis[G::to[i]]) continue;
dfs(G::to[i]);
}
}
unordered_map <int, int> mp;
void solve(vector <int> isl, int n, int m) {
if (isl.size() <= 1) return;
pii st = make_pair(isl.front(), isl.back());
G::fir.fill(0), G::cnt = 1;
for (int i = 1; i <= m; i++) {
G::_add(get <0>(edge[i]), get <1>(edge[i]), get <2>(edge[i]));
G::_add(get <1>(edge[i]), get <0>(edge[i]), get <2>(edge[i]));
}
int ans = Mfl::dinic(st);
vis = 0;
mp[ans]++, dfs(st.fi);
write(ans), puts("");
vector <int> lpr, rpl;
for (auto x : isl) {
if (x == st.fi || x == st.se) continue;
vis[x] ? lpr.push_back(x) : rpl.push_back(x);
}
for (auto x : isl)
write(x), putchar(32);
puts("");
lpr.push_back(st.fi), rpl.push_back(st.se);
solve(lpr, n, m), solve(rpl, n, m);
}
}
int main() {
int n = read(), m = read();
for (int i = 1; i <= m; i++) {
int u = read(), v = read(), w = read();
edge[i] = make_tuple(u, v, w);
}
vector <int> isl; for (int i = 1; i <= n; i++) isl.push_back(i);
Mft::solve(isl, n, m);
write(Mft::mp.size()), puts("");
return 0;
}
标签:cnt,include,int,最小,st,edge,P4123,CQOI2016,dis
From: https://www.cnblogs.com/cxqghzj/p/17894506.html