poj 2195 Going Home【zkw费用流】

题目链接：http://poj.org/problem?id=2195

题意：每个人都走到一个屋子的最小花费。二分图类型

代码：

#include <iostream>  #include <algorithm>  #include <set>  #include <map>  #include <string.h>  #include <queue>  #include <sstream>  #include <stdio.h>  #include <math.h>  #include <stdlib.h>  #include <string>using namespace std;const int MAXN = 1000;const int MAXM = 200000;const int INF = 0x3f3f3f3f;struct Edge{    int to, next, cap, flow, cost;    Edge(int _to = 0, int _next = 0, int _cap = 0, int _flow = 0, int _cost = 0) :        to(_to), next(_next), cap(_cap), flow(_flow), cost(_cost) {}}edge[MAXM];struct ZKW_MinCostMaxFlow{    int head[MAXN], tot;    int cur[MAXN];    int dis[MAXN];    bool vis[MAXN];    int ss, tt, N;//源点、汇点和点的总个数（编号是0~N-1）,不需要额外赋值，调用会直接赋值    int min_cost, max_flow;    void init()    {        tot = 0;        memset(head, -1, sizeof(head));    }    void addedge(int u, int v, int cap, int cost)    {        edge[tot] = Edge(v, head[u], cap, 0, cost);        head[u] = tot++;        edge[tot] = Edge(u, head[v], 0, 0, -cost);        head[v] = tot++;    }    int aug(int u, int flow)    {        if (u == tt)return flow;        vis[u] = true;        for (int i = cur[u];i != -1;i = edge[i].next)        {            int v = edge[i].to;            if (edge[i].cap > edge[i].flow && !vis[v] && dis[u] == dis[v] + edge[i].cost)            {                int tmp = aug(v, min(flow, edge[i].cap - edge[i].flow));                edge[i].flow += tmp;                edge[i ^ 1].flow -= tmp;                cur[u] = i;                if (tmp)return tmp;            }        }        return 0;    }    bool modify_label()    {        int d = INF;        for (int u = 0;u < N;u++)            if (vis[u])                for (int i = head[u];i != -1;i = edge[i].next)                {                    int v = edge[i].to;                    if (edge[i].cap>edge[i].flow && !vis[v])                        d = min(d, dis[v] + edge[i].cost - dis[u]);                }        if (d == INF)return false;        for (int i = 0;i < N;i++)            if (vis[i])            {                vis[i] = false;                dis[i] += d;            }        return true;    }    /*    * 直接调用获取最小费用和最大流    * 输入: start-源点，end-汇点，n-点的总个数（编号从0开始）    * 返回值: pair<int,int> 第一个是最小费用，第二个是最大流    */    pair<int, int> mincostmaxflow(int start, int end, int n)    {        ss = start, tt = end, N = n;        min_cost = max_flow = 0;        for (int i = 0;i < n;i++)dis[i] = 0;        while (1)        {            for (int i = 0;i < n;i++)cur[i] = head[i];            while (1)            {                for (int i = 0;i < n;i++)vis[i] = false;                int tmp = aug(ss, INF);                if (tmp == 0)break;                max_flow += tmp;                min_cost += tmp*dis[ss];            }            if (!modify_label())break;        }        return make_pair(min_cost, max_flow);    }}solve;int n, m;char s[110][110];int main(){    while (~scanf("%d%d", &n, &m))    {        if (n == 0 && m == 0) break;        solve.init();        for (int i = 0;i < n;i++)        {            scanf("%s", s[i]);            for (int j = 0;j < m;j++)                if (s[i][j] == 'H')                    solve.addedge(n*m, i*m + j, 1, 0);                else if (s[i][j] == 'm')                    solve.addedge(i*m + j, n*m + 1, 1, 0);        }        for (int i = 0;i < n*m;i++)            for (int j = 0;j < n*m;j++)            {                int x1 = i / m;                int y1 = i % m;                int x2 = j / m;                int y2 = j % m;                if (s[x1][y1] == 'H' && s[x2][y2] == 'm')                    solve.addedge(i, j, 1, abs(x1 - x2) + abs(y1 - y2));            }        pair<int, int> ans = solve.mincostmaxflow(n * m, n * m + 1, n*m + 2);        printf("%d\n",ans.first);    }    return 0;}