您的位置:首页 > 编程语言 > Go语言

hdu 1533 Going Home 费用流MCMF模板

2015-09-07 16:01 465 查看
题意:n个人和房子,每个人只能进入一个房子,且消费为两者曼哈顿距离,求n个人进入房子的最小花费。

建超源超汇,超源向人连边,超汇向房子连边,然后人向房子连边,跑MCMF即可

#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 pair <int, int> pl;
//typedef __int64 LL;
const int inf = 0x3f3f3f3f;
const int N = 440;
const int st = N-9, ed = N-8;

struct node{
	int v, w, c, nxt;	//c--->cap  w--->cost(dis)
}e[N*N*2];

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

;
int flow, cost;
vector <pl> H, M;

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

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", &n, &m) ) {
        if( n == m && n == 0 )
            break;
        init();
        for( int i = 1; i <= n; ++i ) {
            scanf("%s", mat[i]+1);
            for( int j = 1; j <= m; ++j ) {
                if( mat[i][j] == 'H' )
                    H.push_back( Mp( i, j ) );
                if( mat[i][j] == 'm' )
                    M.push_back( Mp( i, j ) );
            }
        }
        int hh = H.size(), mm = M.size();
        for( int i = 0; i < hh; ++i ) {
            add( st, i, 1, 0 );
            int x1 = H[i].first, y1 = H[i].second;
            for( int j = 0; j < mm; ++j ) {
                int x2 = M[j].first, y2 = M[j].second;
                int DIS =  abs( x1-x2 ) + abs( y1-y2 );
                add( i, j+hh, 1, DIS );
            }
        }
        for( int i = 0; i < mm; ++i )
            add( i+hh, ed, 1, 0 );
        int ans = MCMF();
        printf("%d\n", ans);
	}
	return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: