hdu 4411 Arrest 费用流模板

题意：警察局在0点，里面有k个警察，要将1～n的贼窝一网打尽，这1+n个点都有距离，且要求抓 i 点的贼前保证已经抓光 比i小的贼。警察们最后要回到0点，问满足抓到所以的贼（题目保证可行）最少走的路之和

建边参照yyn http://blog.csdn.net/u013368721/article/details/38781127

#include <bits/stdc++.h>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <time.h>
#include <vector>
#include <cstdio>
#include <string>
#include <iomanip>
///cout << fixed << setprecision(13) << (double) x << endl;
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define ls rt << 1
#define rs rt << 1 | 1
#define pi acos(-1.0)
#define eps 1e-8
#define Mp(a, b) make_pair(a, b)
#define asd puts("asdasdasdasdasdf");
typedef long long ll;
//typedef __int64 LL;
const int inf = 0x3f3f3f3f;
const int M = 100010;
const int N = 220;
const int st = N-9, ed = N-8;

struct node{
	int v, w, c, nxt;	//c--->cap  w--->cost(dis)
}e[M];

int cnt, n, m, k;
int head
;
int dis
;	//费用看成距离
int cap
;		//流量
int cur
;		//当前弧
int vis
;
queue <int> q;
int mat

;
int flow, cost;

void init()
{
	cnt = 0;
	memset( head, -1, sizeof( head ) );
	memset( mat, inf, sizeof( mat ) );
}

void add( int u, int v, int c, int w )
{
	e[cnt].v = v;
	e[cnt].c = c;
	e[cnt].w = w;
	e[cnt].nxt = head[u];
	head[u] = cnt++;

	e[cnt].v = u;
	e[cnt].c = 0;
	e[cnt].w = -w;
	e[cnt].nxt = head[v];
	head[v] = cnt++;
}

bool spfa()
{
	memset( dis, inf, sizeof( dis ) );
	memset( vis, 0, sizeof( vis ) );
	while( !q.empty() )	q.pop();
	cap[st] = inf;
	cur[st] = -1;
	dis[st] = 0;
	q.push( st );
	while( !q.empty() ) {
		int u = q.front();
		q.pop();
		vis[u] = 0;
		for( int i = head[u]; ~i; i = e[i].nxt ) {
			int v = e[i].v, c = e[i].c, w = e[i].w;
			if( c && dis[v] > dis[u] + w ) {
				dis[v] = dis[u] + w;
				cap[v] = min( c, cap[u] );
				cur[v] = i;
				if( !vis[v] ) {
					vis[v] = 1;
					q.push(v);
				}
			}
		}
	}
	if( dis[ed] == inf )
		return 0;
	flow += cap[ed];
	cost += cap[ed] * dis[ed];
	for( int i = cur[ed]; ~i; i = cur[e[i^1].v] ) {
		e[i].c -= cap[ed];
		e[i^1].c += cap[ed];
	}
	return 1;
}

int MCMF()
{
	flow = cost = 0;
	while( spfa() );
	return cost;
}

int main()
{
	while( ~scanf("%d%d%d", &n, &m, &k) ) {
		if( n == m && n == k && n == 0 )
			break;
		init();
		for( int i = 1, u, v, w; i <= m; ++i ) {
			scanf("%d%d%d", &u, &v, &w);
			mat[u][v] = mat[v][u] = min( mat[u][v], w );
		}
		for( int k = 0; k <= n; ++k ) {
			for( int i = 0; i <= n; ++i ) {
				for( int j = 0; j <= n; ++j ) {
					mat[i][j] = min( mat[i][j], mat[i][k]+mat[k][j] );
				}
			}
		}
		for( int i = 1; i <= n; ++i )
			add( i, i+n, 1, -M );		//保证一定要经过i点，但是导致费用为负，在最后加回来就行了。
		for( int i = 1; i <= n; ++i ) {
			add( 0, i, 1, mat[0][i] );	//送一个警察去抓贼
			add( i+n, ed, 1, mat[i][0] );	//回到家
		}
		for( int i = 1; i <= n; ++i ) {
			for( int j = i+1; j <= n; ++j )
				add( i+n, j, 1, mat[i][j] );
		}
		add( st, 0, k, 0 );
		add( 0, ed, k, 0 );
		int ans = MCMF() + n * M;	//加回n*M
		printf("%d\n", ans);
	}
	return 0;
}