题意:
给了两个(x,y)地图..其实是描述的一张图..第一张图说明图中的每个点至多有多少个lizard出去..第二张图说明哪些位置有lizard..现在告诉lizard每次的最远移动距离..而图中任意两点的距离为其曼哈顿距离...问能否让所有的lizard走出地图...
题解:
这题和POJ 3498差不多了...一个点拆成"起点"和"终点"..“起点"到"终点"的流量为最大离开的lizard个数...而超级源点与每个有'L'点的"起点"做边.容量为1..而若一个点可以一步走出去..则其与超级汇点做边..容量为无穷大..然后根据各点能到达的情况做边..跑一次最大流找出最多能跑出去的lizard个数..用总数减去它就是答案..值得注意的是输出要看清..
Program:
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#define MAXN 1005
#define MAXM 2000005
#define oo 1000000007
#define ll long long
using namespace std;
struct Dinic
{
struct node
{
int c,u,v,next;
}edge[MAXM];
int ne,head[MAXN];
int cur[MAXN], ps[MAXN], dep[MAXN];
void initial()
{
ne=2;
memset(head,0,sizeof(head));
}
void addedge(int u, int v,int c)
{
edge[ne].u=u,edge[ne].v=v,edge[ne].c=c,edge[ne].next=head[u];
head[u]=ne++;
edge[ne].u=v,edge[ne].v=u,edge[ne].c=0,edge[ne].next=head[v];
head[v]=ne++;
}
int MaxFlow(int s,int t)
{
int tr, res = 0;
int i,j,k,f,r,top;
while(1)
{
memset(dep, -1, sizeof(dep));
for(f=dep[ps[0]=s]=0,r=1;f!= r;)
for(i=ps[f++],j=head[i];j;j=edge[j].next)
if(edge[j].c&&dep[k=edge[j].v]==-1)
{
dep[k]=dep[i]+1;
ps[r++]=k;
if(k == t){ f=r; break; }
}
if(dep[t]==-1) break;
memcpy(cur,head,sizeof(cur));
i=s,top=0;
while(1)
{
if(i==t)
{
for(tr=oo,k=0;k<top;k++)
if(edge[ps[k]].c<tr)
tr=edge[ps[f=k]].c;
for(k=0;k<top;k++)
{
edge[ps[k]].c-=tr;
edge[ps[k]^1].c+=tr;
}
i=edge[ps[top=f]].u;
res+= tr;
}
for(j=cur[i];cur[i];j=cur[i]=edge[cur[i]].next)
if(edge[j].c && dep[i]+1==dep[edge[j].v]) break;
if(cur[i]) ps[top++]=cur[i],i=edge[cur[i]].v;
else
{
if(!top) break;
dep[i]=-1;
i=edge[ps[--top]].u;
}
}
}
return res;
}
}T;
char A[25][25],B[25][25];
int main()
{
int n,m,d,i,j,ii,jj,s,e,x,sum,ans,C,cases=0;
scanf("%d",&C);
for (cases=1;cases<=C;cases++)
{
scanf("%d%d",&n,&d);
for (i=0;i<n;i++) scanf("%s",A[i]);
m=strlen(A[0]);
for (i=0;i<n;i++) scanf("%s",B[i]);
s=n*m*2+10,e=s+1,T.initial();
sum=0;
for (i=0;i<n;i++)
for (j=0;j<m;j++)
{
x=i*m+j;
if (A[i][j]!='0') T.addedge(x<<1,x<<1|1,A[i][j]-'0');
if (B[i][j]=='L') T.addedge(s,x<<1,1),sum++;
for (ii=0;ii<n;ii++)
for (jj=0;jj<m;jj++)
if (A[ii][jj]!='0' && abs(i-ii)+abs(j-jj)<=d)
T.addedge(x<<1|1,(ii*m+jj)<<1,oo);
if (i-d<0 || i+d>=n || j-d<0 || j+d>=m) T.addedge(x<<1|1,e,oo);
}
ans=sum-T.MaxFlow(s,e);
printf("Case #%d: ",cases);
if (!ans) printf("no");
else printf("%d",ans);
if (ans<=1)
printf(" lizard was left behind.\n");
else printf(" lizards were left behind.\n");
}
return 0;
}
