标签：leapin lizards hdu 2732 最大流
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.. ........ ........
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.
思路：首先将有柱子的点拆成两个，权为承受力，增加超级源点和汇点，‘L’和源点相连，权为1，能一次跳出去的柱子和汇点相连，权为INF，然后能相互到达的柱子之间连边，权为INF。这一题要注意蜥蜴能跳的不只是四个方向，360度任何方向都行，只要两个柱子之间的距离小于d。最后注意输出。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 10005
#define MAXN 50005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
struct Edge{
int u,v,cap,next;
}edge[MAXN];
int n,d;
int dir[4][2]={0,1,0,-1,-1,0,1,0};
int head[maxn],level[maxn],cur[maxn];
int num,cnt,all;
char MP[25][25];
int mp[25][25];
int number[25][25];
void init()
{
cnt=0;
num=0;
all=0;
mem(head,-1);
}
void addedge(int u,int v,int w)
{
edge[num].u=u; edge[num].v=v; edge[num].cap=w; edge[num].next=head[u]; head[u]=num++;
edge[num].u=v; edge[num].v=u; edge[num].cap=0; edge[num].next=head[v]; head[v]=num++;
}
bool bfs(int s,int t)
{
mem(level,-1);
queue<int>Q;
level[s]=0;
Q.push(s);
while (!Q.empty())
{
int u=Q.front(); Q.pop();
for (int i=head[u];i+1;i=edge[i].next)
{
int v=edge[i].v;
if (level[v]==-1&&edge[i].cap>0)
{
level[v]=level[u]+1;
Q.push(v);
}
}
}
return level[t]!=-1;
}
int dfs(int u,int t,int f)
{
if (u==t) return f;
for (int &i=cur[u];i+1;i=edge[i].next)
{
int v=edge[i].v;
if (edge[i].cap>0&&level[v]==level[u]+1)
{
int d=dfs(v,t,min(f,edge[i].cap));
if (d>0)
{
edge[i].cap-=d;
edge[i^1].cap+=d;
return d;
}
}
}
return 0;
}
int dinic(int s,int t,int cnt)
{
int flow=0;
while (bfs(s,t))
{
for (int i=0;i<=cnt;i++) cur[i]=head[i];
int f;
while ((f=dfs(s,t,INF))>0)
flow+=f;
}
return flow;
}
int main()
{
int i,j,k,p,t,cas=0;
sf(t);
while (t--)
{
init();
int len;
sff(n,d);
for (i=0;i<n;i++)
{
scanf("%s",MP[i]);
len=strlen(MP[i]);
for (j=0;j<len;j++)
{
mp[i][j]=MP[i][j]-'0';
number[i][j]=++cnt;
}
}
mem(MP,0);
for (i=0;i<n;i++)
scanf("%s",MP[i]);
int t=2*cnt+1;
for (i=0;i<n;i++)
{
for (j=0;j<len;j++)
{
if (mp[i][j])
{
if (i<d||j<d||n-i<=d||len-j<=d)
addedge(number[i][j]+cnt,t,INF);
addedge(number[i][j],number[i][j]+cnt,mp[i][j]); //拆点
for (k=0;k<n;k++)
{
for (p=0;p<len;p++)
{
int dx=abs(i-k);
int dy=abs(j-p);
double mm=sqrt(dx*dx*1.0+dy*dy*1.0);
if (mm>d) continue;
addedge(number[i][j]+cnt,number[k][p],INF);
}
}
}
if (MP[i][j]=='L')
{
addedge(0,number[i][j],1);
all++;
}
}
}
int s=dinic(0,t,t+1);
int ans=all-s;
if (!ans)
printf("Case #%d: no lizard was left behind.\n",++cas);
else if (ans==1)
printf("Case #%d: 1 lizard was left behind.\n",++cas);
else
printf("Case #%d: %d lizards were left behind.\n",++cas,ans);
}
return 0;
}
Leapin' Lizards (hdu 2732 最大流)
标签：leapin lizards hdu 2732 最大流
原文地址：http://blog.csdn.net/u014422052/article/details/45116963
