标签:string ios space algo zoj names ati bzoj list
*****我很想爆粗但是要文明好气噢
我是真的翻大车了
这题我一看这不是费用流吗zz
然后感觉强连通直接记一下出度入度不久行了吗,然后码完自信1WA
回来改费用流建图烦的要死,结果是n打成m。。。浪费时间没有收获
做法都会啊拆点然后向四周出点连边,开始方向费用0其他为1
#include<cstdio> #include<iostream> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> using namespace std; const int inf=999999999; struct node { int x,y,c,d,next,other; }a[210000];int len,last[510]; void ins(int x,int y,int c,int d) { int k1,k2; len++;k1=len; a[len].x=x;a[len].y=y;a[len].c=c;a[len].d=d; a[len].next=last[x];last[x]=len; len++;k2=len; a[len].x=y;a[len].y=x;a[len].c=0;a[len].d=-d; a[len].next=last[y];last[y]=len; a[k1].other=k2; a[k2].other=k1; } int ans,st,ed; int list[510];bool v[510]; int d[510],pre[510],cc[510]; bool spfa() { memset(d,63,sizeof(d));d[st]=0; memset(cc,0,sizeof(cc));cc[st]=inf; memset(v,false,sizeof(v));v[st]=true; int head=1,tail=2;list[1]=st; while(head!=tail) { int x=list[head]; for(int k=last[x];k;k=a[k].next) { int y=a[k].y; if(d[y]>d[x]+a[k].d&&a[k].c>0) { d[y]=d[x]+a[k].d; cc[y]=min(cc[x],a[k].c); pre[y]=k; if(v[y]==false) { v[y]=true; list[tail]=y; tail++;if(tail==505)tail=1; } } } v[x]=false; head++;if(head==505)head=1; } if(d[ed]>=inf)return false; else { ans+=cc[ed]*d[ed]; int y=ed; while(pre[y]!=0) { int k=pre[y]; a[k].c-=cc[ed]; a[a[k].other].c+=cc[ed]; y=a[k].x; } return true; } } int n,m; int point(int x,int y){return (x-1)*m+y;} int U[20][20],D[20][20],L[20][20],R[20][20]; char ss[20]; int main() { scanf("%d%d",&n,&m);st=2*n*m+1,ed=2*n*m+2; for(int i=1;i<=n;i++) { scanf("%s",ss+1); for(int j=1;j<=m;j++) { U[i][j]=1;D[i][j]=1;L[i][j]=1;R[i][j]=1; if(ss[j]==‘U‘)U[i][j]=0; else if(ss[j]==‘D‘)D[i][j]=0; else if(ss[j]==‘L‘)L[i][j]=0; else if(ss[j]==‘R‘)R[i][j]=0; } } for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { if(i==1)ins(point(i,j),point(n,j)+n*m,1,U[i][j]); else ins(point(i,j),point(i-1,j)+n*m,1,U[i][j]); if(i==n)ins(point(i,j),point(1,j)+n*m,1,D[i][j]); else ins(point(i,j),point(i+1,j)+n*m,1,D[i][j]); if(j==1)ins(point(i,j),point(i,m)+n*m,1,L[i][j]); else ins(point(i,j),point(i,j-1)+n*m,1,L[i][j]); if(j==m)ins(point(i,j),point(i,1)+n*m,1,R[i][j]); else ins(point(i,j),point(i,j+1)+n*m,1,R[i][j]); ins(st,point(i,j),1,0); ins(point(i,j)+n*m,ed,1,0); } ans=0; while(spfa()==true); printf("%d\n",ans); return 0; }
标签:string ios space algo zoj names ati bzoj list
原文地址:https://www.cnblogs.com/AKCqhzdy/p/8946612.html