【JSOI2009】游戏

题面

https://www.lydsy.com/JudgeOnline/problem.php?id=1443

题解

二分图博弈问题，找一定在匹配上的点，即求可能割边。

一定在最大流上的边竟然是可能割边，因为如果一条边一定在最大流上，把他删了最大流肯定会减少，也就是最小割会减少，所以是可能割边。

这是“最小”和“最大”性质的矛盾之处导致的。

#include<stack>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#define ri register int
#define N 10500
#define S 0
#define T (wyr+1)
using namespace std;

const int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int n,m,bh[105][105],wyr;
char g[105][105];

struct network_flow {
  vector<int> ed[N],to,w;
  int d[N],cur[N];
  int bel[N],dfn[N],low[N],cc,cnt;
  bool ins[N]; stack<int> stk;
  
  void add_edge(int u,int v,int aw) {
    to.push_back(v); w.push_back(aw); ed[u].push_back(to.size()-1);
    to.push_back(u); w.push_back(0);  ed[v].push_back(to.size()-1);
  }
  bool bfs() {
    queue<int> q;
    memset(d,0x3f,sizeof(d));
    d[S]=0; q.push(S);
    while (!q.empty()) {
      int x=q.front(); q.pop();
      for (ri i=0;i<ed[x].size();i++) {
        int e=ed[x][i];
        if (w[e] && d[x]+1<d[to[e]]) {
          d[to[e]]=d[x]+1;
          q.push(to[e]);
        }
      }
    }
    return d[T]<1000000007;
  }
  int dfs(int x,int lim) {
    if (x==T || !lim) return lim;
    int sum=0;
    for (ri &i=cur[x];i<ed[x].size();i++) {
      int e=ed[x][i];
      if (w[e] && d[x]+1==d[to[e]]) {
        int f=dfs(to[e],min(lim,w[e]));
        w[e]-=f; w[1^e]+=f;
        sum+=f; lim-=f;
        if (!lim) return sum;
      }
    }
    return sum;
  }
  int dinic() {
    int ret=0;
    while (bfs()) {
      memset(cur,0,sizeof(cur));
      ret+=dfs(S,n*m+1);
    }
    return ret;
  }
  void tarjan(int x) {
    dfn[x]=low[x]=++cc;
    ins[x]=1; stk.push(x);
    for (ri i=0;i<ed[x].size();i++) {
      int e=ed[x][i];
      if (!w[e]) continue;
      if (dfn[to[e]]) {
        if (ins[to[e]]) low[x]=min(low[x],dfn[to[e]]);
      }
      else {
        tarjan(to[e]);
        low[x]=min(low[x],low[to[e]]);
      }
    }
    if (dfn[x]==low[x]) {
      int t; ++cnt;
      do {
        t=stk.top(); stk.pop(); ins[t]=0;
        bel[t]=cnt;
      }
      while (t!=x);
    }
  }
} G;

bool check() {
  for (ri i=S;i<=T;i++) if (!G.dfn[i]) G.tarjan(i);
  for (ri i=1;i<=n;i++)
    for (ri j=1;j<=m;j++) if (bh[i][j]) {
      if ((i+j)%2==1) {
        if (G.bel[bh[i][j]]!=G.bel[S] && !G.w[2*bh[i][j]-2]) ; else return 1;
      }
      else {
        if (G.bel[bh[i][j]]!=G.bel[T] && !G.w[2*bh[i][j]-2]) ; else return 1;
      }
    }
  return 0;
}

int main() {
  scanf("%d %d
",&n,&m);
  memset(g,'#',sizeof(g));
  for (ri i=1;i<=n;i++) scanf("%s",g[i]+1);
  wyr=0;
  for (ri i=1;i<=n;i++) 
    for (ri j=1;j<=m;j++) if (g[i][j]=='.') bh[i][j]=++wyr;
      
  for (ri i=1;i<=n;i++)
    for (ri j=1;j<=m;j++) if (bh[i][j]) {
      if ((i+j)%2==1) G.add_edge(S,bh[i][j],1); else G.add_edge(bh[i][j],T,1);
    }
  for (ri i=1;i<=n;i++)
    for (ri j=1;j<=m;j++) if (bh[i][j] && (i+j)%2==1) {
      for (ri k=0;k<4;k++) {
        int nx=i+dx[k],ny=j+dy[k];
        if (bh[nx][ny]) G.add_edge(bh[i][j],bh[nx][ny],1);
      }
    }
  G.dinic();
  puts(check()?"WIN":"LOST");
  for (ri i=1;i<=n;i++) 
    for (ri j=1;j<=m;j++) if (bh[i][j]) {
      if ((i+j)%2==1) {
        if (G.bel[bh[i][j]]!=G.bel[S] && !G.w[2*bh[i][j]-2]) ; else printf("%d %d
",i,j);
      }
      else {
        if (G.bel[bh[i][j]]!=G.bel[T] && !G.w[2*bh[i][j]-2]) ; else printf("%d %d
",i,j);
      }
    }
  return 0;
}