题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2732
| Problem Description Your platoon of wandering lizards has entered a strange room in the labyrinth you are exploring. As you are looking around for hidden treasures, one of the rookies steps on an innocent-looking stone and the room's floor suddenly disappears! Each lizard in your platoon is left standing on a fragile-looking pillar, and a fire begins to rage below... Leave no lizard behind! Get as many lizards as possible out of the room, and report the number of casualties. Input The input file will begin with a line containing a single integer representing the number of test cases, which is at most 25. Each test case will begin with a line containing a single positive integer n representing the number of rows in the map, followed by a single non-negative integer d representing the maximum leaping distance for the lizards. Two maps will follow, each as a map of characters with one row per line. The first map will contain a digit (0-3) in each position representing the number of jumps the pillar in that position will sustain before collapsing (0 means there is no pillar there). The second map will follow, with an 'L' for every position where a lizard is on the pillar and a '.' for every empty pillar. There will never be a lizard on a position where there is no pillar.Each input map is guaranteed to be a rectangle of size n x m, where 1 ≤ n ≤ 20 and 1 ≤ m ≤ 20. The leaping distance is Output For each input case, print a single line containing the number of lizards that could not escape. The format should follow the samples provided below. Sample Input Sample Output |
题目大意：T组测试数据，每组测试数据输入 n（n行的地图）没有m，需要自己找，再给一个D，一只蜥蜴行走的最远距离， 再给出两个图，分别是某个位置所能行走的次数，后面的那个图是蜥蜴的初始位置，看蜥蜴们能不能逃出这个地图（出地图边界）
很明显的拆点，一个平台分成两半，自己与自己联通的是流量限制，与其他联通的是INF；
下面的图给的是与 源 的连线；
能直接跳出来的与 汇 直接连；
一般难度拆点题：我感觉用上弧优化会更快一点，但是看时间，似乎不用弧优化也可以过，但我没试
ac;
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<map>
//#include<set>
#include<deque>
#include<queue>
#include<stack>
#include<bitset>
#include<string>
#include<fstream>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define mod 998244353
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b)
#define clean(a,b) memset(a,b,sizeof(a))// 水印
//std::ios::sync_with_stdio(false);
struct dot{
int x,y;
}d[100010];
struct node{
int v,w,cost,nxt;
node(int _v=0,int _w=0,int _nxt=0):
v(_v),w(_w),nxt(_nxt){}
}edge[500005<<1];
int head[500005],cur[500005],ecnt;
int dis[500005];
int n,m,s,t;
int D,sum;
void intt()
{
clean(head,-1);
clean(cur,-1);
ecnt=0;
s=0,t=801;
}
void add(int u,int v,int w)
{
edge[ecnt]=node(v,w,head[u]);
head[u]=ecnt++;
edge[ecnt]=node(u,0,head[v]);
head[v]=ecnt++;
}
/*---上面的是板子，不用动---*/
bool bfs()
{
clean(dis,-1);
dis[s]=0;
queue<int> que;
que.push(s);
while(que.size())
{
int u=que.front();
que.pop();
if(u==t)
return 1;
for(int i=head[u];i+1;i=edge[i].nxt)
{
int temp=edge[i].v;
if(dis[temp]==-1&&edge[i].w>0)
{
dis[temp]=dis[u]+1;
que.push(temp);
}
}
}
return 0;
}
int dfs(int u,int low)
{
if(u==t||low==0)
return low;
int res=0;
for(int &i=cur[u];i+1;i=edge[i].nxt)
{
int temp=edge[i].v;
if(dis[temp]==dis[u]+1&&edge[i].w>0)
{
int f=dfs(temp,min(low-res,edge[i].w));
edge[i].w-=f;
edge[i^1].w+=f;
res=res+f;
if(res==low)
break;
}
}
return res;
}
int dinic()
{
int ans=0,res;
while(bfs())
{
for(int i=s;i<=t;++i)
cur[i]=head[i];
ans=ans+dfs(s,INF);
}
return ans;
}
int main()
{
int T,Cas=1;
cin>>T;
while(T--)
{
intt();
cin>>n>>D;
sum=0;
string str;
for(int i=1;i<=n;++i)//读入次数
{
cin>>str;
if(i==1)
m=str.size();
int len=m;
for(int j=0;j<len;++j)
{
if(str[j]-'0'>0)
{
int u=(i-1)*m+j+1;
add(u,u+400,str[j]-'0');//拆点
if(i-D<=0||i+D>n||j-D<0||j+D>=m)
add(u+400,t,INF);//直接跳出地图
else
{
for(int k=1;k<=n;++k)
{
for(int h=0;h<m;++h)
{
int u2=(k-1)*m+h+1;
if(u==u2)
continue;
if(abs(i-k)+abs(j-h)<=D)
add(u+400,u2,INF);
}
}
}
}
}
}
for(int i=1;i<=n;++i)//读入柱子
{
cin>>str;
for(int j=0;j<m;++j)
{
int u=(i-1)*m+j+1;
if(str[j]=='L')
{
++sum;//加一只蜥蜴
add(s,u,1);
}
}
}
// for(int i=s;i<=t;++i)//图没错
// {
// cout<<i<<" : ";
// for(int j=head[i];j+1;j=edge[j].nxt)
// cout<<edge[j].v<<" "<<edge[j].w<<" ->>";
// cout<<endl;
// }
int ans=sum-dinic();
//cout<<ans<<" "<<sum-ans<<endl;
if(ans==0)
cout<<"Case #"<<Cas++<<": no lizard was left behind."<<endl;
else if(ans==1)
cout<<"Case #"<<Cas++<<": 1 lizard was left behind."<<endl;
else
cout<<"Case #"<<Cas++<<": "<<ans<<" lizards were left behind."<<endl;
}
}