POJ - 2195 Going Home

１．题面

http://poj.org/problem?id=2195

２．题意

最小费用最大流裸题，在一张格点地图上，从一个格子移动到相邻的格子代价为１，每个格子可以站无数个人，求让所有人都走进屋子的最小代价，每个屋子只能待一个人，每个人只能进一个屋子。

３．思路

添加一个公共源点，向每个人连一条边，费用为０，容量为１，

添加一个公共汇点，由每个屋子往汇点连一条边，费用为０，容量为１．

再从人往屋子连边，费用为人与屋子之间的距离。

再求最小费用流就可以了。

４．代码

/*****************************************************************
> File Name: cpp_acm.cpp
> Author: Uncle_Sugar
> Mail: uncle_sugar@qq.com
> Created Time: Mon 03 Oct 2016 19:59:11 CST
*****************************************************************/
# include <cstdio>
# include <cstring>
# include <cctype>
# include <cmath>
# include <cstdlib>
# include <climits>
# include <iostream>
# include <iomanip>
# include <set>
# include <map>
# include <vector>
# include <stack>
# include <queue>
# include <algorithm>
using namespace std;

# define rep(i,a,b) for (i=a;i<=b;i++)
# define rrep(i,a,b) for (i=b;i>=a;i--)
# define mset(aim, val) memset(aim, val, sizeof(aim))

struct QuickIO{
QuickIO(){const int SZ = 1<<20;
setvbuf(stdin ,new char[SZ],_IOFBF,SZ);
setvbuf(stdout,new char[SZ],_IOFBF,SZ);
}				//*From programcaicai*//
}QIO;

template<class T>void PrintArray(T* first,T* last,char delim=' '){
for (;first!=last;first++) cout << *first << (first+1==last?'\n':delim);
}

/*
1.see the size of the input data before you select your algorithm
2.cin&cout is not recommended in ACM/ICPC
3.pay attention to the size you defined, for instance the size of edge is double the size of vertex
*/

const int debug = 1;
const int size  = 10 + 100;
const int INF = INT_MAX>>1;
typedef long long ll;
const int MAXN = 100 + 100*100;
const int MAXM = 100000;

struct Edge{
int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数,节点编号从0~N-1

void init(int n){
N = n; tol = 0;
memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int cap,int cost){
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}

bool spfa(int s,int t){
queue<int>q;
for(int i = 0;i < N;i++){
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while(!q.empty()){
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u]; i != -1;i = edge[i].next){
int v = edge[i].to;
if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost ){
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if(!vis[v]){
vis[v] = true;
q.push(v);
}
}
}
}
if(pre[t] == -1)return false;
else return true;
}
//返回的是最大流,cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost){
int flow = 0;
cost = 0;
while(spfa(s,t)){
//# cout << "hello" << endl;
int Min = INF;
for(int i = pre[t];i != -1;i = pre[edge[i^1].to]){
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for(int i = pre[t];i != -1;i = pre[edge[i^1].to]){
edge[i].flow += Min;
edge[i^1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
return flow;
}

typedef pair<int, int> pir;

vector<pir> man, house;

int dist(const pir& p1, const pir& p2){
return abs(p1.first - p2.first) + abs(p1.second - p2.second);
}

int main()
{
/*std::ios::sync_with_stdio(false);cin.tie(0);*/
int n, m;
while (~scanf("%d%d", &n, &m)){
if (n==0&&m==0) break;
man.clear();house.clear();
char str[size];
for (int i = 0; i < n; i++){
scanf("%s", str);
for (int j = 0; j < m; j++){
switch(str[j]){
case 'm':man.push_back(pir(i, j));break;
case 'H':house.push_back(pir(i, j));break;
}
}
}
int sz = man.size();
init(sz + sz + 2);
for (int i = 1; i <= sz; i++){
addedge(0, i, 1, 0);
addedge(sz + i, sz + sz + 1, 1, 0);
}
for (int i = 0; i < sz; i++){
for (int j = 0; j < sz; j++){
addedge(i + 1, sz + j + 1, 1, dist(man[i], house[j]));
}
}
int cost;
//# cout << "hello" << endl;
int ans = minCostMaxflow(0, sz + sz + 1, cost);
printf("%d\n", cost);
}
return 0;
}