4 6 1 0 0 1 0 0 0 1 1 0 0 0 2 0 0 0 0 0 0 2 0 1 1 0
Case 1: 4
题意:一个矩阵,1表示羊,2表示狼,然后让把羊和狼隔开,问需要最小的栅栏数。
思路:要把羊和狼隔开,就是去最小割把他们分开,每个格子和周围四个格子建流量为1的边,s向狼建INF的边,羊向t建INF的边。
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 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;
const int MAXM = 200010;
const int MAXN = 43000;
int n,m,N;
int mp[205][205];
struct Edge
{
int to,next,cap,flow;
}edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init(int n)
{
tol=0;N=n;
memset(head,-1,sizeof(head));
}
//加边,单向图三个参数,双向图四个参数
void addedge(int u,int v,int w,int rw=0)
{
edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u];
edge[tol].flow=0; head[u]=tol++;
edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v];
edge[tol].flow=0; head[v]=tol++;
}
//输入参数:起点,终点,点的总数
//点的编号没有影响,只要输入点的总数
int sap(int start,int end,int N)
{
memset(gap,0,sizeof(gap));
memset(dep,0,sizeof(dep));
memcpy(cur,head,sizeof(head));
int u=start;
pre[u]=-1;
gap[0]=N;
int ans=0;
while (dep[start]<N)
{
if (u==end)
{
int Min=INF;
for (int i=pre[u];i!=-1;i=pre[edge[i^1].to])
if (Min>edge[i].cap-edge[i].flow)
Min=edge[i].cap-edge[i].flow;
for (int i=pre[u];i!=-1;i=pre[edge[i^1].to])
{
edge[i].flow+=Min;
edge[i^1].flow-=Min;
}
u=start;
ans+=Min;
continue;
}
bool flag=false;
int v;
for (int i=cur[u];i!=-1;i=edge[i].next)
{
v=edge[i].to;
if (edge[i].cap-edge[i].flow && dep[v]+1==dep[u])
{
flag=true;
cur[u]=pre[v]=i;
break;
}
}
if (flag)
{
u=v;
continue;
}
int Min=N;
for (int i=head[u];i!=-1;i=edge[i].next)
if (edge[i].cap-edge[i].flow && dep[edge[i].to]<Min)
{
Min=dep[edge[i].to];
cur[u]=i;
}
gap[dep[u]]--;
if (!gap[dep[u]]) return ans;
dep[u]=Min+1;
gap[dep[u]]++;
if (u!=start) u=edge[pre[u]^1].to;
}
return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/asus1/Desktop/IN.txt","r",stdin);
#endif
int i,j,cas=0;
while (~sff(n,m))
{
for (i=0;i<n;i++)
for (j=0;j<m;j++)
sf(mp[i][j]);
init(n*m+2);
int s=n*m,t=s+1;
for (i=0;i<n;i++)
{
for (j=0;j<m;j++)
{
if (i>0)
addedge(i*m+j,(i-1)*m+j,1);
if (j>0)
addedge(i*m+j,i*m+j-1,1);
if (i<(n-1))
addedge(i*m+j,(i+1)*m+j,1);
if (j<(m-1))
addedge(i*m+j,i*m+j+1,1);
if (mp[i][j]==1)
addedge(i*m+j,t,INF);
if (mp[i][j]==2)
addedge(s,i*m+j,INF);
}
}
pf("Case %d:\n",++cas);
pf("%d\n",sap(s,t,N));
}
return 0;
}
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Pleasant sheep and big big wolf (hdu 3046 最小割)
原文:http://blog.csdn.net/u014422052/article/details/46848179