【题解】Catering World Finals 2015 上下界费用流

Prelude

传送到Codeforces：0.0

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Solution

板子题，在这里贴个板子。

这题面是smg？题面中有说每个点只能经过一次吗？是我瞎了吗？

因为这WA on test 27一个小时，烦死了，浪费时间。

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Code

#include <cstring>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <cassert>

using namespace std;
const int MAXN = 1000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
int _w;

int read() {
	int x;
	_w = scanf( "%d", &x );
	return x;
}

namespace MCMF {
	struct Edge {
		int u, v, c, f, w;
		Edge() {}
		Edge( int u, int v, int c, int f, int w ):
			u(u), v(v), c(c), f(f), w(w) {}
	};

	int n, m, s, t;
	int head[MAXN], nxt[MAXM];
	Edge edge[MAXM];

	void init( int _n ) {
		n = _n, m = 0;
		for( int i = 0; i < n; ++i )
			head[i] = -1;
	}
	void adde( int u, int v, int c, int w ) {
		edge[m] = Edge(u, v, c, 0, w);
		nxt[m] = head[u], head[u] = m++;
		edge[m] = Edge(v, u, 0, 0, -w);
		nxt[m] = head[v], head[v] = m++;
	}

	queue<int> q;
	int dis[MAXN], inq[MAXN], res[MAXN], from[MAXN];

	bool spfa() {
		for( int i = 0; i < n; ++i )
			dis[i] = INF, inq[i] = 0;
		dis[s] = 0, inq[s] = 1, q.push(s), res[s] = INF;
		while( !q.empty() ) {
			int u = q.front(); q.pop();
			inq[u] = 0;
			for( int i = head[u]; ~i; i = nxt[i] ) {
				const Edge &e = edge[i];
				if( e.c > e.f && dis[u] + e.w < dis[e.v] ) {
					dis[e.v] = dis[u] + e.w;
					from[e.v] = i;
					res[e.v] = min( res[u], e.c-e.f );
					if( !inq[e.v] )
						inq[e.v] = 1, q.push(e.v);
				}
			}
		}
		return dis[t] != INF;
	}
	void augment() {
		int f = res[t], u = t;
		while( u != s ) {
			int i = from[u];
			edge[i].f += f;
			edge[i^1].f -= f;
			u = edge[i].u;
		}
	}
	int solve( int _s, int _t ) {
		s = _s, t = _t;
		int cost = 0;
		while( spfa() ) {
			cost += res[t] * dis[t];
			augment();
		}
		return cost;
	}
}

int n, k, g[MAXN][MAXN];
int s, t, ss, tt, nid;
int in[MAXN], out[MAXN];

void adde( int u, int v, int l, int r, int w ) {
	MCMF::adde(ss, v, l, 0);
	MCMF::adde(u, tt, l, 0);
	MCMF::adde(u, v, r-l, w);
}

void solve() {
	s = nid++, t = nid++, ss = nid++, tt = nid++;
	for( int i = 1; i <= n; ++i )
		in[i] = nid++, out[i] = nid++;
	MCMF::init(nid);
	adde(t, s, 0, INF, 0);
	adde(s, out[1], 0, k, 0);
	for( int i = 2; i <= n; ++i ) {
		adde(out[i], t, 0, INF, 0);
		adde(in[i], out[i], 1, 1, 0); // 每个点只能经过一次
	}
	for( int i = 1; i <= n; ++i )
		for( int j = i+1; j <= n; ++j )
			adde(out[i], in[j], 0, INF, g[i][j]);
	printf( "%d\n", MCMF::solve(ss, tt) );
}

int main() {
	n = read()+1, k = read();
	for( int i = 1; i <= n; ++i )
		for( int j = i+1; j <= n; ++j )
			g[i][j] = read();
	solve();
	return 0;
}