Leapin' Lizards(hdu 2732)

Leapin' Lizards

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2080    Accepted Submission(s): 857

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.
The pillars in the room are aligned as a grid, with each pillar one unit away from the pillars to its east, west, north and south. Pillars at the edge of the grid are one unit away from the edge of the room (safety). Not all pillars necessarily have a lizard. A lizard is able to leap onto any unoccupied pillar that is within d units of his current one. A lizard standing on a pillar within leaping distance of the edge of the room may always leap to safety... but there's a catch: each pillar becomes weakened after each jump, and will soon collapse and no longer be usable by other lizards. Leaping onto a pillar does not cause it to weaken or collapse; only leaping off of it causes it to weaken and eventually collapse. Only one lizard may be on a pillar at any given time.

 

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
always 1 ≤ d ≤ 3.

 

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.

 思路：最大流，将每个点拆分成两个，一个为入流点一个为出流点，两点间建立流量为这个点处柱子的高度。

然后建立一个总输出点0，和总汇点，还有在两个距离不超过d的点之间建立流量为无穷的边，0和每个L的输入点建立流为1的边。最后跑下dinic。用所有的L-最大流即可。

#include<stdio.h>
#include<algorithm>
#include<stdlib.h>
#include<iostream>
#include<string.h>
#include<vector>
#include<queue>
using namespace std;
struct node
{
        int to;
        int cap;
        int rev;
};
char str[30][30];
char  str2[30][30];
int id[30][30];
int level[100000];
vector<node>vec[100000];
const int N=1e8;
int iter[100000];
void add(int from,int to,int cap);
void bfs(int s);
int dfs(int s,int t,int f);
int max_flow(int s,int t);
int  main(void)
{
        int i,j,k;
        scanf("%d",&k);
        int s;
        int p,q;
        for(s=1; s<=k; s++)
        {
                scanf("%d %d ",&p,&q);
                for(i=0; i<100000; i++)
                        vec[i].clear();
                for(i=0; i<p; i++)
                {
                        scanf("%s",str[i]);
                }
                for(i=0; i<p; i++)
                {
                        scanf("%s",str2[i]);
                }
                int l=strlen(str[0]);
                int ans=1;
                for(i=0; i<p; i++)
                {
                        for(j=0; j<l; j++)
                        {
                                id[i][j]=ans++;
                        }
                }
                ans-=1;
                int x,y;
                for(i=0; i<p; i++)
                {
                        for(j=0; j<l; j++)
                        {
                                if(str[i][j]>'0')
                                        add(id[i][j],id[i][j]+ans,str[i][j]-'0');
                        }
                }
                for(i=0; i<p; i++)
                {
                        for(j=0; j<l; j++)
                        {
                                for(x=0; x<p; x++)
                                {
                                        for(y=0; y<l; y++)
                                        {
                                                if(i!=x||j!=y)
                                                {
                                                        if(abs(x-i)+abs(j-y)<=q)
                                                        {
                                                                add(id[i][j]+ans,id[x][y],N);
                                                        }
                                                }
                                        }
                                }
                        }
                }
                for(i=0; i<p; i++)
                {
                        for(j=0; j<l; j++)
                        {
                                int cc=abs(i+1);
                                int dd=abs(p-i);
                                cc=min(cc,dd);
                                int kk=abs(j+1);
                                int vv=abs(l-j);
                                vv=min(kk,vv);
                                if((vv<=q||cc<=q))
                                {
                                        add(id[i][j]+ans,2*ans+1,N);
                                }
                        }
                }
                int fuck=0;
                for(i=0; i<p; i++)
                {
                        for(j=0; j<l; j++)
                        {
                                if(str2[i][j]=='L')
                                {
                                        add(0,id[i][j],1);
                                        fuck++;
                                }
                        }
                }
                int sum=max_flow(0,2*ans+1);
                printf("Case #%d: ",s);
                if(fuck-sum==1)
                printf("%d lizard was left behind.
",fuck-sum);
                else if(fuck-sum==0)  printf("no lizard was left behind.
");
                else  printf("%d lizards were left behind.
",fuck-sum);
        }
}
void add(int from,int to,int cap)
{
        node nn;
        nn.to=to;
        nn.cap=cap;
        nn.rev=vec[to].size();
        vec[from].push_back(nn);
        nn.to=from;
        nn.cap=0;
        nn.rev=vec[from].size()-1;
        vec[to].push_back(nn);
}
void bfs(int s)
{
        queue<int>que;
        memset(level,-1,sizeof(level));
        level[s]=0;
        que.push(s);
        while(!que.empty())
        {
                int v=que.front();
                que.pop();
                int i;
                for(i=0; i<vec[v].size(); i++)
                {
                        node e=vec[v][i];
                        if(level[e.to]==-1&&e.cap>0)
                        {
                                level[e.to]=level[v]+1;
                                que.push(e.to);
                        }
                }
        }
}
int dfs(int s,int t,int f)
{
        if(s==t)
                return f;
        for(int &i=iter[s]; i<vec[s].size(); i++)
        {
                node &e=vec[s][i];
                if(level[e.to]>level[s]&&e.cap>0)
                {
                        int r=dfs(e.to,t,min(e.cap,f));
                        if(r>0)
                        {
                                e.cap-=r;
                                vec[e.to][e.rev].cap+=r;
                                return r;
                        }
                }
        }
        return 0;
}
int max_flow(int s,int t)
{
        int flow=0;
        for(;;)
        {
                bfs(s);
                if(level[t]<0)return flow;
                memset(iter,0,sizeof(iter));
                int f;
                while((f=dfs(s,t,N))>0)
                {
                        flow+=f;
                }
        }
}