「雅礼集训 2017 Day2」棋盘游戏

祝各位圣诞后快乐(逃)

题目传送门

分析:

首先棋盘上的路径构成的图是一张二分图

那么对于一个二分图,先求出最大匹配,先手如果走到关键匹配点,只要后手顺着匹配边走,由于不再会出现增广路径,所以走到最后先手就必败

所以Alice只要到非关键匹配点,Bob便一定会走到关键匹配点,然后Alice便必胜

于是求一下那些点是非关键匹配点就好了

真难想233

我好菜233

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>

#define maxn 500005
#define maxm 105
#define INF 0x3f3f3f3f

using namespace std;

inline long long getint()
{
    long long num=0,flag=1;char c;
    while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
    while(c>='0'&&c<='9')num=num*10+c-48,c=getchar();
    return num*flag;
}

int n,m;
int S,T;
int fir[maxn],nxt[maxn],to[maxn],cap[maxn],cnt;
int h[maxn],vis[maxn];
char s[maxm][maxm];
int ans[maxm][maxm];

inline void newnode(int u,int v,int w)
{to[++cnt]=v,nxt[cnt]=fir[u],fir[u]=cnt,cap[cnt]=w;}
inline void insert(int u,int v,int w)
{newnode(u,v,w),newnode(v,u,0);}

inline bool bfs()
{
    memset(h,-1,sizeof h);h[S]=0;
    queue<int>Q;Q.push(S);
    while(!Q.empty())
    {
        int u=Q.front();Q.pop();
        for(int i=fir[u];i;i=nxt[i])
            if(!~h[to[i]]&&cap[i])h[to[i]]=h[u]+1,Q.push(to[i]);
    }
    return ~h[T];
}

inline int aug(int u,int flow)
{
    if(u==T||!flow)return flow;
    int used=0;
    for(int i=fir[u];i;i=nxt[i])
        if(cap[i]&&h[to[i]]==h[u]+1)
        {
            int delta=aug(to[i],min(flow-used,cap[i]));
            cap[i]-=delta,cap[i^1]+=delta,used+=delta;
            if(used==flow)return flow;
        }
    if(!used)h[u]=-1;
    return used;
}

inline int dinic()
{
    int num=0;
    while(bfs())num+=aug(S,INF);
    return num;
}

inline void dfs(int u)
{
    vis[u]=1;
    for(int i=fir[u];i;i=nxt[i])if(!vis[to[i]]&&cap[i])dfs(to[i]);
}

inline void dfs2(int u)
{
    vis[u]=1;
    for(int i=fir[u];i;i=nxt[i])if(!vis[to[i]]&&cap[i^1])dfs2(to[i]);
}

inline int getid(int i,int j){return i*m+j-m;}

int main()
{
    n=getint(),m=getint();int num=0;
    for(int i=1;i<=n;i++)scanf("%s",s[i]+1);
    S=n*m+1,T=S+1;cnt=1;
    for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
    {
        if(((i+j)&1)&&s[i][j]=='.')
        {
            insert(S,getid(i,j),1);
            if(s[i-1][j]=='.')insert(getid(i,j),getid(i-1,j),INF);
            if(s[i+1][j]=='.')insert(getid(i,j),getid(i+1,j),INF);
            if(s[i][j-1]=='.')insert(getid(i,j),getid(i,j-1),INF);
            if(s[i][j+1]=='.')insert(getid(i,j),getid(i,j+1),INF);
        }
        if(!((i+j)&1)&&s[i][j]=='.')insert(getid(i,j),T,1);
    }
    dinic();
    dfs(S);
    for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
        if(((i+j)&1)&&s[i][j]=='.'&&vis[getid(i,j)])ans[i][j]=1,num++;
    memset(vis,0,sizeof vis);
    dfs2(T);
    for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
        if(!((i+j)&1)&&s[i][j]=='.'&&vis[getid(i,j)])ans[i][j]=1,num++;
    printf("%d\n",num);
    for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
        if(ans[i][j])printf("%d %d\n",i,j);
}
View Code

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转载自www.cnblogs.com/Darknesses/p/12104277.html