标签:style class blog http tar ext
题目链接:http://61.187.179.132/JudgeOnline/problem.php?id=2668
题意:有一个n行m列的黑白棋盘,你每次可以交换两个相邻格子中的棋子,最终达到目标状态。要求第i行第j列的格子只能参与m[i,j]次交换。
思路: 我们将1看做要移动的数字,将0看做空白。那么若1在始末状态个数不同则无解;如某个格子始末状态均有1则这个格子的1对结果无影响,可以将其都置为0。将每个格子拆为为个点p0,p1,p2:
(1)若格子初始为1,则连边:<s,p0,1,0>,<p1,p0,m[i][j]/2,0)>,<p0,p2,(m[i][j]+1)/2,0>;
(2)若格子末状态为0,则连边:<p0,t,1,0>,<p1,p0,(m[i][j]+1)/2,0>,<p0,p2,m[i][j]/2,0>;
(3)始末都是空白,则连边:<p1,p0,m[i][j]/2,0>,<p0,p2,m[i][j]/2,0>;
(4)相邻格子x和y连边<px2,py1,INF,0>。
struct node
{
int u,v,next,cost,cap;
};
node edges[N];
int head[N],e;
void add(int u,int v,int cap,int cost)
{
edges[e].u=u;
edges[e].v=v;
edges[e].cap=cap;
edges[e].cost=cost;
edges[e].next=head[u];
head[u]=e++;
}
void Add(int u,int v,int cap,int cost)
{
add(u,v,cap,cost);
add(v,u,0,-cost);
}
int pre[N],F[N],C[N],visit[N];
int SPFA(int s,int t,int n)
{
int i;
for(i=0;i<=n;i++) F[i]=0,C[i]=INF,visit[i]=0;
queue<int> Q;
Q.push(s); F[s]=INF; C[s]=0;
int u,v,cost,cap;
while(!Q.empty())
{
u=Q.front();
Q.pop();
visit[u]=0;
for(i=head[u];i!=-1;i=edges[i].next)
{
if(edges[i].cap>0)
{
v=edges[i].v;
cost=edges[i].cost;
cap=edges[i].cap;
if(C[v]>C[u]+cost)
{
C[v]=C[u]+cost;
F[v]=min(F[u],cap);
pre[v]=i;
if(!visit[v]) visit[v]=1,Q.push(v);
}
}
}
}
return F[t];
}
char a[25][25],b[25][25],c[25][25];
int d[25][25][3];
int dx[]={-1,-1,-1,0,1,1,1,0};
int dy[]={-1,0,1,1,1,0,-1,-1};
int n,m,s,t,cnt;
int ans;
int MCMF(int s,int t,int n)
{
int i,x,temp,M=0;
while(temp=SPFA(s,t,n))
{
M+=temp;
for(i=t;i!=s;i=edges[pre[i]].u)
{
x=pre[i];
ans+=edges[x].cost*temp;
edges[x].cap-=temp;
edges[x^1].cap+=temp;
}
}
return M==cnt;
}
int main()
{
RD(n,m);
int i,j;
FOR1(i,n) RD(a[i]+1);
FOR1(i,n) RD(b[i]+1);
FOR1(i,n) RD(c[i]+1);
int k=0;
FOR1(i,n) FOR1(j,m)
{
a[i][j]-=‘0‘;
b[i][j]-=‘0‘;
c[i][j]-=‘0‘;
d[i][j][0]=++k;
d[i][j][1]=++k;
d[i][j][2]=++k;
if(a[i][j]&&b[i][j]) a[i][j]=0,b[i][j]=0;
}
s=0; t=++k;
clr(head,-1);
cnt=0;
int x,y,p=0;
FOR1(i,n) FOR1(j,m)
{
if(a[i][j])
{
cnt++;
Add(s,d[i][j][0],1,0);
Add(d[i][j][1],d[i][j][0],c[i][j]/2,0);
Add(d[i][j][0],d[i][j][2],(c[i][j]+1)/2,0);
}
else if(b[i][j])
{
p++;
Add(d[i][j][0],t,1,0);
Add(d[i][j][1],d[i][j][0],(c[i][j]+1)/2,0);
Add(d[i][j][0],d[i][j][2],c[i][j]/2,0);
}
else
{
Add(d[i][j][1],d[i][j][0],c[i][j]/2,0);
Add(d[i][j][0],d[i][j][2],c[i][j]/2,0);
}
FOR0(k,8)
{
x=i+dx[k];
y=j+dy[k];
if(x>=1&&x<=n&&y>=1&&y<=m)
{
Add(d[i][j][2],d[x][y][1],INF,1);
}
}
}
if(cnt!=p||!MCMF(s,t,t+1)) puts("-1");
else PR(ans);
}
BZOJ 2668 交换棋子(费用流),布布扣,bubuko.com
标签:style class blog http tar ext
原文地址:http://www.cnblogs.com/jianglangcaijin/p/3799793.html