Leapin' Lizards（经典建图，最大流）

Leapin' Lizards

http://acm.hdu.edu.cn/showproblem.php?pid=2732

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4180    Accepted Submission(s): 1670

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.

 

Sample Input
4
3 1
1111
1111
1111
LLLL
LLLL
LLLL
3 2
00000
01110
00000
.....
.LLL.
.....
3 1
00000
01110
00000
.....
.LLL.
.....
5 2
00000000
02000000
00321100
02000000
00000000
........
........
..LLLL..
........
........

Sample Output
Case #1: 2 lizards were left behind.
Case #2: no lizard was left behind.
Case #3: 3 lizards were left behind.
Case #4: 1 lizard was left behind.

 

 

一道水题做一天。。。。真菜

拆点跑最大流就好，距离是曼哈顿距离

 

#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#define maxn 200005
#define MAXN 200005
#define mem(a,b) memset(a,b,sizeof(a))
const int N=200005;
const int M=200005;
const int INF=0x3f3f3f3f;
using namespace std;
int n;
struct Edge{
    int v,next;
    int cap,flow;
}edge[MAXN*20];//注意这里要开的够大。。不然WA在这里真的想骂人。。问题是还不报RE。。
int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
int cnt=0;//实际存储总边数
void isap_init()
{
    cnt=0;
    memset(pre,-1,sizeof(pre));
}
void isap_add(int u,int v,int w)//加边
{
    edge[cnt].v=v;
    edge[cnt].cap=w;
    edge[cnt].flow=0;
    edge[cnt].next=pre[u];
    pre[u]=cnt++;
}
void add(int u,int v,int w){
    isap_add(u,v,w);
    isap_add(v,u,0);
}
bool bfs(int s,int t)//其实这个bfs可以融合到下面的迭代里，但是好像是时间要长
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]=1;
    dep[t]=0;
    queue<int>q;
    while(!q.empty())
    q.pop();
    q.push(t);//从汇点开始反向建层次图
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=pre[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].v;
            if(dep[v]==-1&&edge[i^1].cap>edge[i^1].flow)//注意是从汇点反向bfs，但应该判断正向弧的余量
            {
                dep[v]=dep[u]+1;
                gap[dep[v]]++;
                q.push(v);
                //if(v==sp)//感觉这两句优化加了一般没错，但是有的题可能会错，所以还是注释出来，到时候视情况而定
                //break;
            }
        }
    }
    return dep[s]!=-1;
}
int isap(int s,int t)
{
    if(!bfs(s,t))
    return 0;
    memcpy(cur,pre,sizeof(pre));
    //for(int i=1;i<=n;i++)
    //cout<<"cur "<<cur[i]<<endl;
    int u=s;
    path[u]=-1;
    int ans=0;
    while(dep[s]<n)//迭代寻找增广路,n为节点数
    {
        if(u==t)
        {
            int f=INF;
            for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增广路
                f=min(f,edge[i].cap-edge[i].flow);
            for(int i=path[u];i!=-1;i=path[edge[i^1].v])
            {
                edge[i].flow+=f;
                edge[i^1].flow-=f;
            }
            ans+=f;
            u=s;
            continue;
        }
        bool flag=false;
        int v;
        for(int i=cur[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].v;
            if(dep[v]+1==dep[u]&&edge[i].cap-edge[i].flow)
            {
                cur[u]=path[v]=i;//当前弧优化
                flag=true;
                break;
            }
        }
        if(flag)
        {
            u=v;
            continue;
        }
        int x=n;
        if(!(--gap[dep[u]]))return ans;//gap优化
        for(int i=pre[u];i!=-1;i=edge[i].next)
        {
            if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x)
            {
                x=dep[edge[i].v];
                cur[u]=i;//常数优化
            }
        }
        dep[u]=x+1;
        gap[dep[u]]++;
        if(u!=s)//当前点没有增广路则后退一个点
        u=edge[path[u]^1].v;
     }
     return ans;
}

string mp[505];
string book[505];
int dir[4][2]={0,1,1,0,0,-1,-1,0};
int main(){
    std::ios::sync_with_stdio(false);
    int m,s,t;
    int f,d;
    int T;
    cin>>T;
    for(int co=1;co<=T;co++){
        cin>>n>>d;
        for(int i=0;i<n;i++) cin>>book[i];
        for(int i=0;i<n;i++) cin>>mp[i];
        isap_init();
        m=mp[0].length();
        int sum=0;
        s=n*m*2,t=n*m*2+1;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(mp[i][j]=='L'){
                    add(s,i*m+j,1);
                    sum++;
                }
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(book[i][j]!='0'){
                    if(i<d||j<d||i+d>=n||j+d>=m){
                        add(m*n+i*m+j,t,book[i][j]-'0');
                    }
                    add(i*m+j,n*m+i*m+j,book[i][j]-'0');
                    for(int ii=-d;ii<=d;ii++){
                        for(int jj=-d;jj<=d;jj++){
                            int xx=i+ii,yy=j+jj;
                            if(xx>=0&&xx<n&&yy>=0&&yy<m){
                                if(book[xx][yy]!='0'&&(ii||jj)){
                                    if(abs(ii)+abs(jj)<=d){
                                        add(n*m+i*m+j,xx*m+yy,INF);
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
        n=n*m*2+2;
        int ans=isap(s,t);
        ans=sum-ans;
        cout<<"Case #"<<co<<": ";
        if(!ans) cout<<"no lizard was left behind."<<endl;
        else if(ans==1) cout<<"1 lizard was left behind."<<endl;
        else cout<<ans<<" lizards were left behind."<<endl;
    }
}