POJ训练计划2195_Going Home(网络流/费用流)

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解题报告

题目传送门

思路：

bfs建图跑一下费用流就行。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define inf 0x3f3f3f3f

using namespace std;
struct E {
    int v,cost,cap,next;
} edge[100000];
int head[1000],cnt,dis[1000],pre[1000],vis[1000],s,t,mmap[110][110],h,w,f[1100],cost,flow;
struct N {
    int x,y,step;
};
void add(int u,int v,int cost,int cap) {
    edge[cnt].v=v;
    edge[cnt].cost=cost;
    edge[cnt].cap=cap;
    edge[cnt].next=head[u];
    head[u]=cnt++;

    edge[cnt].v=u;
    edge[cnt].cost=-cost;
    edge[cnt].cap=0;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}
int dx[]= {-1,0,1,0};
int dy[]= {0,1,0,-1};
void bfs(int x,int y) {
    queue<N>q;
    int v[110][110];
    memset(v,0,sizeof(v));
    N next,now;
    now.x=x;
    now.y=y;
    now.step=0;
    v[x][y]=1;
    q.push(now);
    while(!q.empty()) {
        now=q.front();
        q.pop();
        if(mmap[now.x][now.y]>=101) {
            add(mmap[x][y],mmap[now.x][now.y],now.step,1);
        }
        for(int i=0; i<4; i++) {
            next.x=now.x+dx[i];
            next.y=now.y+dy[i];
            if(next.x>=0&&next.x<h&&next.y>=0&&next.y<w&&!v[next.x][next.y]) {
                next.step=now.step+1;
                v[next.x][next.y]=1;
                q.push(next);
            }
        }
    }
}
int _spfa() {
    for(int i=0; i<=201; i++) {
        dis[i]=inf;
        vis[i]=pre[i]=f[i]=0;
    }
    dis[s]=0;
    vis[s]=1;
    pre[s]=-1;
    f[s]=inf;
    queue<int>Q;
    Q.push(s);
    while(!Q.empty()) {
        int u=Q.front();
        Q.pop();
        vis[u]=0;
        for(int i=head[u]; i!=-1; i=edge[i].next) {
            int v=edge[i].v;
            if(edge[i].cap&&dis[v]>dis[u]+edge[i].cost) {
                pre[v]=i;
                dis[v]=dis[u]+edge[i].cost;
                f[v]=min(f[u],edge[i].cap);
                if(!vis[v]) {
                    vis[v]=1;
                    Q.push(v);
                }
            }
        }
    }
    if(dis[t]==inf)return 0;
    flow+=f[t];
    cost+=f[t]*dis[t];
    for(int i=pre[t]; i!=-1; i=pre[edge[i^1].v]) {
        edge[i].cap-=f[t];
        edge[i^1].cap+=f[t];
    }
    return 1;
}
void mcmf() {
    cost=flow=0;
    while(_spfa());
    printf("%d\n",cost);
}
int main() {
    int i,j,n,m;
    char str[1000];
    while(~scanf("%d%d",&h,&w)) {
        n=101,m=1;
        s=0,t=201;
        if(!h&&!w)break;
        memset(edge,0,sizeof(edge));
        memset(head,-1,sizeof(head));
        cnt=0;
        for(i=0; i<h; i++) {
            scanf("%s",str);
            for(j=0; j<w; j++) {
                if(str[j]=='.')
                    mmap[i][j]=-1;
                else if(str[j]=='H')
                    mmap[i][j]=n++;
                else if(str[j]=='m')
                    mmap[i][j]=m++;
            }
        }
        for(i=0; i<h; i++) {
            for(j=0; j<w; j++) {
                if(mmap[i][j]>=1&&mmap[i][j]<=100) {
                    add(0,mmap[i][j],0,1);
                    bfs(i,j);
                }
                if(mmap[i][j]>=101)
                    add(mmap[i][j],t,0,1);
            }
        }
        mcmf();
    }
    return 0;
}

Description

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man. 

Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a ‘.‘ means an empty space, an ‘H‘ represents a house on that point, and am ‘m‘ indicates there is a little man on that point. 

You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.

Input

Output

Sample Input

2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0

Sample Output

2
10
28

Source

POJ训练计划2195_Going Home(网络流/费用流)

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原文地址：http://blog.csdn.net/juncoder/article/details/38780109