hdoj 1533 && poj 2195 Going Home

题目连接：Going Home

题目大意：现在有一张图，有n个房间和n个人，现在每个人移动的代价是1，只能上下左右移动，不能有相同的人在一个房间，问所有的人都进入房间的最小花费是多少

题目思路：这种有限制而且还有花费的，我们不妨转化为最小费用最大流，我们用一个超级源点连接所有的人，容量为1，单位花费为0，然后用一个超级汇点被所有的房间连接，容量为1，单位花费为0，然后每个人和所有房间相连，容量为1，单位花费为这个人和这个房间的距离，然后直接跑最小费用最大流就好了

#include <map>#include <set>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;const int INF = 0x3f3f3f3f;const int MaxNode = 205;const int MaxEdge = 40005;struct Edge{    int to,vol,cost,next;}Edges[MaxEdge];int Pre[MaxNode],Path[MaxNode],Dist[MaxNode],Head[MaxNode],EdgeCount;void addedge(int u, int v, int vol, int cost){    Edges[EdgeCount].to = v;    Edges[EdgeCount].vol = vol;    Edges[EdgeCount].cost = cost;    Edges[EdgeCount].next = Head[u];    Head[u] = EdgeCount++;    Edges[EdgeCount].to = u;    Edges[EdgeCount].vol = 0;    Edges[EdgeCount].cost = -cost;    Edges[EdgeCount].next = Head[v];    Head[v] = EdgeCount++;}bool Spfa(int s, int t){    memset(Dist, 0x3f3f3f3f, sizeof(Dist));    memset(Pre, -1, sizeof(Pre));    Dist[s] = 0;    queue<int>Q;    Q.push(s);    while (!Q.empty()){        int u = Q.front();        Q.pop();        for (int e = Head[u]; e != -1; e = Edges[e].next){            int v = Edges[e].to;            if (Edges[e].vol > 0 && Dist[v] > Dist[u] + Edges[e].cost){                Dist[v] = Dist[u] + Edges[e].cost;                Pre[v] = u;                Path[v] = e;                Q.push(v);            }        }    }    return Pre[t] != -1;}int MinCostFlow(int s, int t){    int cost = 0;    int max_flow = 0;    int u, v, e;    while (Spfa(s, t)){        int f = INF;        for (u = t; u != s; u = Pre[u]){            f = min(f, Edges[Path[u]].vol);        }        for (u = t; u != s; u = Pre[u]){            e = Path[u];            Edges[e].vol -= f;            Edges[e^1].vol += f;        }        max_flow += f;        cost += f*Dist[t];    }    return cost;}void init(){    memset(Head,-1,sizeof(Head));    EdgeCount = 0;}char gMap[MaxNode][MaxNode];vector<pair<int, int> > gHousVec;vector<pair<int, int> > gManVec;int main(){    int n,m;    char mp[MaxNode][MaxNode];    while(~scanf("%d%d",&n,&m)&&n+m){        init();        gHousVec.clear();        gManVec.clear();        for(int i = 1;i <= n;i++){            getchar();            for(int j = 1;j <= m;j++){                scanf("%c",&mp[i][j]);                if(mp[i][j] == 'm') gManVec.push_back(pair<int,int>(i,j));                else if(mp[i][j] == 'H') gHousVec.push_back(pair<int,int>(i,j));            }        }        int u = gHousVec.size();        for (int i = 1; i <= u; i++) addedge(0, i, 1, 0);        for (int i = 0; i < u; i++){            for (int j = 0; j < u; j++){                int value = abs(gManVec[i].first - gHousVec[j].first) + abs(gManVec[i].second - gHousVec[j].second);                addedge(i + 1, j + u + 1, 1, value);            }        }        for (int i = 1; i <= u; i++) addedge(u + i, 2 * u + 1, 1, 0);        int cost = MinCostFlow(0,2*u+1);        printf("%d\n",cost);    }    return 0;}