HDU 1533 最小费用最大流（模板）

Posted 2020-06-19 Donser

http://acm.hdu.edu.cn/showproblem.php?pid=1533

这道题直接用了模板

题意：要构建一个二分图，家对应人，连线的权值就是最短距离，求最小费用

要注意void init(int n) 这个函数一定要写

一开始忘记写这个WA了好几发

还有这个题很容易T掉，赋值建图要简化，一开始构建成网络流那种图一直T

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#define oo 0x13131313
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN=400;
const int MAXM=200000;
struct Edge
{
    int to,next,cap,flow,cost;
} edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;
void init(int n)
{
    N = n;
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
    edge[tol].to = v;
    edge[tol].cap = cap;
    edge[tol].cost = cost;
    edge[tol].flow = 0;
    edge[tol].next = head[u];
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = 0;
    edge[tol].cost = -cost;
    edge[tol].flow = 0;
    edge[tol].next = head[v];
    head[v] = tol++;
}
bool spfa(int s,int t)
{
    queue<int>q;
    for(int i = 0; i < N; i++)
    {
        dis[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    dis[s] = 0;
    vis[s] = true;
    q.push(s);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int i = head[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(edge[i].cap > edge[i].flow &&
                    dis[v] > dis[u] +edge[i].cost)
            {
                dis[v] = dis[u] + edge[i].cost;
                pre[v] = i;
                if(!vis[v])
                {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t] == -1)return false;
    else return true;
}
int minCostMaxflow(int s,int t,int &cost)
{
    int flow = 0;
    cost = 0;
    while(spfa(s,t))
    {
        int Min = INF;
        for(int i = pre[t]; i != -1 ; i = pre[edge[i^1].to])
        {
            if(Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
        }
        for(int i = pre[t]; i != -1; i = pre[edge[i^1].to])
        {
            edge[i].flow += Min;
            edge[i^1].flow -= Min;
            cost += edge[i].cost*Min;
        }
        flow += Min;
    }
    return flow;
}
struct Home
{
    int x,y;
} H[MAXN],P[MAXN];
int main()
{
    int totH,totP;
    int NN,MM;
    while(cin>>NN>>MM&&NN&&MM)
    {
        getchar();
        init(MAXN);
        char c;
        totH=0;
        totP=0;
        for(int i=1; i<=NN; i++)
        {
            for(int j=1; j<=MM; j++)
            {
                scanf("%c",&c);
                if(c==‘H‘) totH++,H[totH].x=i,H[totH].y=j;
                else if(c==‘m‘) totP++,P[totP].x=i,P[totP].y=j;
            }
            getchar();
        }

        int ANS=0;
        int NNN=totP+totH;
        for(int i=1; i<=totP; i++)
            for(int j=1; j<=totH; j++)
            {
                int t=abs(P[i].x-H[j].x)+abs(P[i].y-H[j].y);
                addedge(i,j+totP,1,t);
            }
        for(int i=1; i<=totP; i++)
            addedge(NNN+1,i,1,0);
        for(int i=totP+1; i<=NNN; i++)
            addedge(i,NNN+2,1,0);
        minCostMaxflow(NNN+1,NNN+2,ANS);
        printf("%d\n",ANS);
    }
    return 0;
}