POJ2195：Going Home——题解

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

题目大意：

有些人和房子，一个人只能进一个房子，人走到房子的路程即为代价。

求所有人走到房子后的最小代价。

——————————————————

bfs处理每个人到每个房的最短路之后就是裸的费用流了，不解释。

#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
#include<algorithm>
#include<cctype>
using namespace std;
typedef long long ll;
const int INF=1e9;
const int N=10010;
const int M=22000;
inline int read(){
    int X=0,w=0;char ch=0;
    while(!isdigit(ch)){w|=ch=='-';ch=getchar();}
    while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
    return w?-X:X;
}
struct node{
    int nxt;
    int to;
    int w;
    int b;
}edge[M];
int head[N],cnt=-1;
void add(int u,int v,int w,int b){
    cnt++;
    edge[cnt].to=v;
    edge[cnt].w=w;
    edge[cnt].b=b;
    edge[cnt].nxt=head[u];
    head[u]=cnt;
    return;
}
int dis[N];
bool vis[N];
inline bool spfa(int s,int t,int n){
    deque<int>q;
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)dis[i]=INF;
    dis[t]=0;q.push_back(t);vis[t]=1;
    while(!q.empty()){
    int u=q.front();
    q.pop_front();vis[u]=0;
    for(int i=head[u];i!=-1;i=edge[i].nxt){
        int v=edge[i].to;
        int b=edge[i].b;
        if(edge[i^1].w&&dis[v]>dis[u]-b){
        dis[v]=dis[u]-b;
        if(!vis[v]){
            vis[v]=1;
            if(!q.empty()&&dis[v]<dis[q.front()]){
            q.push_front(v);
            }else{
            q.push_back(v);
            }
        }
        }
    }
    }
    return dis[s]<INF;
}
int ans=0;
int dfs(int u,int flow,int m){
    if(u==m){
    vis[m]=1;
    return flow;
    }
    int res=0,delta;
    vis[u]=1;
    for(int e=head[u];e!=-1;e=edge[e].nxt){
        int v=edge[e].to;
    int b=edge[e].b;
        if(!vis[v]&&edge[e].w&&dis[u]-b==dis[v]){
            delta=dfs(v,min(edge[e].w,flow-res),m);
            if(delta){
                edge[e].w-=delta;
                edge[e^1].w+=delta;
                res+=delta;
        ans+=delta*b;
                if(res==flow)break;
            }
        }
    }
    return res;
}
inline int costflow(int S,int T,int n){
    while(spfa(S,T,n)){
    do{
        memset(vis,0,sizeof(vis));
        dfs(S,INF,T);
    }while(vis[T]);
    }
    return ans;
}
int num1=0,num2=0;
int mp[101][101];
int pos[101][101];
int pdis[101][101];
bool walk[101][101];
int dx[4]={0,-1,0,1};
int dy[4]={1,0,-1,0};
void bfs(int xx,int yy,int n,int m){
    queue<int>q1,q2,q3,q4;
    memset(walk,0,sizeof(walk));
    memset(pdis,127,sizeof(pdis));
    pdis[xx][yy]=0;
    q1.push(xx);q2.push(yy);
    walk[xx][yy]=1;
    while(!q1.empty()){
    int x=q1.front(),y=q2.front();
    q1.pop(),q2.pop();
    if(mp[x][y]>0){
        q3.push(x);q4.push(y);
    }
    for(int i=0;i<4;i++){
        int nx=x+dx[i],ny=y+dy[i];
        if(nx<=0||ny<=0||nx>n||ny>m||walk[nx][ny])continue;
        walk[nx][ny]=1;
        pdis[nx][ny]=pdis[x][y]+1;
        q1.push(nx);q2.push(ny);
    }
    }
    while(!q3.empty()){
    int nx=q3.front(),ny=q4.front();
    q3.pop();q4.pop();
    add(pos[xx][yy],mp[nx][ny]+num1,1,pdis[nx][ny]);
    add(mp[nx][ny]+num1,pos[xx][yy],0,-pdis[nx][ny]);
    }
    return;
}
void restart(){
    memset(head,-1,sizeof(head));
    memset(mp,0,sizeof(mp));
    memset(pos,0,sizeof(pos));
    cnt=-1;
    ans=0;
    num1=0;
    num2=0;
    return;
}
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF&&n&&m){
    restart();
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
        char ch;
        cin>>ch;
        if(ch=='m'){
            num1++;
            mp[i][j]=-1;
            pos[i][j]=num1;
        }else if(ch=='H'){
            num2++;
            mp[i][j]=num2;
        }
        }
    }
    int S=num1+num2+1,T=S+1;
    for(int i=1;i<=num1;i++){
        add(S,i,1,0);
        add(i,S,0,0);
    }
    for(int i=1;i<=num2;i++){
        add(i+num1,T,1,0);
        add(T,i+num1,0,0);
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
        if(mp[i][j]==-1){
            bfs(i,j,n,m);
        }
        }
    }
    printf("%d
",costflow(S,T,T));
    }
    return 0;
}