HDU-1533 Going Home

Going Home

题意：给你一张图，图上有若干个人和若干个屋子，现在要使的这若干个人都进到屋子里，并且一个屋子只能进一个人，求总步数最小。

题解：最小费用流。将图转化成边的关系，然后求解。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
const int INF = 0x3f3f3f3f;
const LL mod =  (int)1e9+7;
const int N = 50005;
char str[205][205];
int px[N], py[N];
int head[N], to[N], ct[N], w[N], nt[N];
int d[N], vis[N];
int pre[N], id[N];
int n, m, s, t, tot;

void add(int u, int v, int val, int cost){
    to[tot] = v;
    ct[tot] = cost;
    w[tot] = val;
    nt[tot] = head[u];
    head[u] = tot++;
}

void get_G(){
    int p = 0, z = 0;
    s = 0;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            if(str[i][j] == 'H'){
                ++p;
                px[p] = i;
                py[p] = j;
            }
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            if(str[i][j] == 'm'){
                z++;
                add(s,z+p,1,0);
                add(z+p,s,0,0);
                for(int k = 1; k <= p; k++){
                    add(z+p,k,1,abs(i-px[k])+abs(j-py[k]));
                    add(k,z+p,0,-(abs(i-px[k])+abs(j-py[k])));
                }
            }
    t = p + z + 1;
    for(int i = 1; i <= p; i++){
        add(i,t,1,0);
        add(t,i,0,0);
    }
}
void init(){
    memset(head,-1,sizeof(head));
    tot = 0;
}

int spfa(){
    queue<int> q;
    memset(d, INF, sizeof(d));
    memset(vis, 0, sizeof(vis));
    memset(pre, -1, sizeof(vis));
    d[s] = 0;
    q.push(s);
    while(!q.empty()){
        int u = q.front(); q.pop();
        vis[u] = 0;
        for(int i = head[u]; ~i; i = nt[i]){
            if(w[i] > 0 && d[to[i]] > d[u] + ct[i]){
                d[to[i]] = d[u] + ct[i];
                pre[to[i]] = u;
                id[to[i]] = i;
                if(!vis[to[i]]){
                    vis[to[i]] = 1;
                    q.push(to[i]);
                }
            }
        }

    }
    return d[t] < INF;
}

int MaxFlow(){
    int Mi = INF;
    int sum = 0;
    while(spfa()){
        Mi = INF;
        for(int i = t; i != s; i = pre[i])
            Mi = min(Mi, w[id[i]]);
        for(int i = t; i != s; i = pre[i]){
            w[id[i]] -= Mi;
            w[id[i]^1] += Mi;
        }
        sum += d[t];
    }
    return sum;
}

int main(){
    while(~scanf("%d%d", &n, &m), n+m){
        init();
        for(int i = 1; i <= n; i++)
            scanf("%s", str[i]+1);
        get_G();
        printf("%d
", MaxFlow());
    }
    return 0;
}

第一次写KM费用流，跑的特别慢，希望下次能改进。