Oil Skimming HDU - 4185

匈牙利

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char mp[660][660];
int g[660][660];
int relate[660][660];
int match[660];
int vis[660];
int k,n;
int tmp;
int dfs(int u){
    int i;
    for(i = 0; i < tmp; i++){
        if(g[u][i] && !vis[i]){
            vis[i] = 1;
            if(!match[i] || dfs(match[i])){
                match[i] = u;
                return 1;
            }
        }
    }
    return 0;
}
 
int hungry(){
    int i;
    int cnt = 0;
    memset(match,0,sizeof(match));
    for(i = 0; i < tmp; i++){
        memset(vis,0,sizeof(vis));
        if(dfs(i))
            cnt++;
    }
    return cnt;
}
int main(){
    cin >> k;
    int cas = 0;
    while(k--){
        cin >> n;
        getchar();
        memset(relate,0,sizeof(relate));
        memset(g,0,sizeof(g));
        int i,j;
        tmp = 0;
        for(i = 0; i < n; i++){
            cin >> mp[i];
            getchar();
            for(j = 0; j < n; j++){
                if(mp[i][j] == '#')
                    relate[i][j] = tmp++;
            }
        }
        for(i = 0; i < n; i++){
            for(j = 0; j < n; j++){
                if(mp[i][j] != '#') continue;
                if(mp[i-1][j] == '#') g[relate[i][j]][relate[i-1][j]] = 1;
                if(mp[i+1][j] == '#') g[relate[i][j]][relate[i+1][j]] = 1;
                if(mp[i][j-1] == '#') g[relate[i][j]][relate[i][j-1]] = 1;
                if(mp[i][j+1] == '#') g[relate[i][j]][relate[i][j+1]] = 1;
            }
        }
        int ans = hungry();
        printf("Case %d: %d\n",++cas,ans/2);//除以2是因为每个点都作为起点看了一遍,即正反都看了一遍所以是真正答案的两倍
    }
    return 0;
}

HK

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=600*600+10;
const int M=600*600*2+10;
const int INF=0x7f7f7f7f;
const int DX[]={0,0,-1,1};
const int DY[]={-1,1,0,0};
int e[M],ne[M],h[M],idx;
void addedge(int a,int b)
{
	e[idx]=b;
	ne[idx]=h[a];
	h[a]=idx++;
}
void init()
{
    idx=0;
    memset(h,-1,sizeof(h));
}
 
int uN;
int Mx[N],My[N];
int dx[N],dy[N];
int dis;
bool used[N];
bool SearchP()
{
    queue<int> q;
    memset(dx,-1,sizeof(dx));
    memset(dy,-1,sizeof(dy));
    for(int i=1;i<=uN;i++)
        if(Mx[i]==-1)
        {
            q.push(i);
            dx[i]=0;
        }
    dis=INF;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        if(dx[u]>dis) break;
        for(int i=h[u];i!=-1;i=ne[i])
        {
            int v=e[i];
            if(dy[v]==-1)
            {
                dy[v]=dx[u]+1;
                if(My[v]==-1) dis=dy[v];
                else
                {
                    dx[My[v]]=dy[v]+1;
                    q.push(My[v]);
                }
            }
        }
    }
    return dis!=INF;
}
bool dfs(int u)
{
    for(int i=h[u];i!=-1;i=ne[i])
    {
        int v=e[i];
        if(!used[v]&&dy[v]==dx[u]+1)
        {
            used[v]=true;
            if(My[v]!=-1&&dy[v]==dis) continue;
            if(My[v]==-1||dfs(My[v]))
            {
                My[v]=u;
                Mx[u]=v;
                return true;
            }
        }
    }
    return false;
}
int MaxMatch()
{
    int res=0;
    memset(Mx,-1,sizeof(Mx));
    memset(My,-1,sizeof(My));
    while(SearchP())
    {
        memset(used,false,sizeof(used));
        for(int i=1;i<=uN;i++)
            if(Mx[i]==-1&&dfs(i)) res++;
    }
    return res;
}
char g[610][610];
int id[610][610];
int main()
{
    int n,T,kase=0;
    scanf("%d",&T);
    while(T--)
    {
        init();
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%s",g[i]+1);
        memset(id,-1,sizeof(id));
        int cnt=1;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(g[i][j]=='#'&&(i+j)%2==0)
                    id[i][j]=cnt++;
        uN=cnt-1;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(g[i][j]=='#'&&(i+j)%2==1)
                {
                    int t=cnt++;
                    for(int k=0;k<4;k++)
                    {
                        int ni=i+DX[k];
                        int nj=j+DY[k];
                        if(id[ni][nj]!=-1) addedge(id[ni][nj],t);
                    }
                }
        printf("Case %d: %d\n",++kase,MaxMatch());
    }
    return 0;
}

posted @ 2020-03-05 22:04  晴屿  阅读(91)  评论(0编辑  收藏  举报