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POJ 2195 - Going Home 更新最小费用最大流模板..整合成结构体...

Ewall_熊猫 2022-08-12 阅读 45


           Ps...我发现现在这个代码怎么优化还是没有以前写的那个快..以前那个还是用的临接矩阵..真是蛋疼....不过也够用了...


Program:

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<queue>
#include<stack>
#include<set>
#include<time.h>
#include<map>
#include<algorithm>
#define ll long long
#define eps 1e-5
#define oo 1000000007
#define pi acos(-1.0)
#define MAXN 250
#define MAXM 500005
using namespace std;
char S[110][110];
int H[110],M[110];
struct MCMF
{
struct node
{
int x,y,c,v,next;
}line[MAXM];
int Lnum,_next[MAXN],pre[MAXN],dis[MAXN],flow,cost;
bool inqueue[MAXN];
void initial(int n)
{
Lnum=-1;
for (int i=0;i<=n;i++) _next[i]=-1;
}
void addline(int x,int y,int c,int v)
{
line[++Lnum].next=_next[x],_next[x]=Lnum;
line[Lnum].x=x,line[Lnum].y=y,line[Lnum].c=c,line[Lnum].v=v;
line[++Lnum].next=_next[y],_next[y]=Lnum;
line[Lnum].x=y,line[Lnum].y=x,line[Lnum].c=0,line[Lnum].v=-v;
}
bool SPFA(int s,int e)
{
int x,k,y;
queue<int> Q;
while (!Q.empty()) Q.pop();
memset(dis,0x7f,sizeof(dis));
memset(inqueue,false,sizeof(inqueue));
Q.push(s);
dis[s]=0,pre[s]=-1;
while (!Q.empty())
{
x=Q.front(),Q.pop(),inqueue[x]=false;
for (k=_next[x];k!=-1;k=line[k].next)
if (line[k].c)
{
y=line[k].y;
if (dis[y]>dis[x]+line[k].v)
{
dis[y]=dis[x]+line[k].v;
pre[y]=k;
if (!inqueue[y])
{
inqueue[y]=true;
Q.push(y);
}
}
}
}
if (dis[e]>oo) return false;
flow=oo,cost=0;
for (k=pre[e];k!=-1;k=pre[line[k].x])
flow=min(flow,line[k].c),cost+=line[k].v;
cost*=flow;
for (k=pre[e];k!=-1;k=pre[line[k].x])
line[k].c-=flow,line[k^1].c+=flow;
return true;
}
int MinCostMaxFlow(int s,int e)
{
int Aflow=0,Acost=0;
while (SPFA(s,e))
{
Aflow+=flow;
Acost+=cost;
}
return Acost;
}
}T;
int main()
{
int R,C,i,j,k,n,m,s,e;
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
while (~scanf("%d%d",&R,&C) && (R || C))
{
for (i=0;i<R;i++) scanf("%s",S[i]);
n=m=0;
for (i=0;i<R;i++)
for (j=0;j<C;j++)
{
if (S[i][j]=='m') M[++n]=i*C+j;
if (S[i][j]=='H') H[++m]=i*C+j;
}
s=n+m+1,e=s+1,T.initial(e);
for (i=1;i<=n;i++) T.addline(s,i,1,0);
for (i=1;i<=m;i++) T.addline(i+n,e,1,0);
for (i=1;i<=n;i++)
for (j=1;j<=m;j++)
T.addline(i,j+n,1,abs(M[i]/C-H[j]/C)+abs(M[i]%C-H[j]%C));
printf("%d\n",T.MinCostMaxFlow(s,e));
}
return 0;
}



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