hrbust 1976 Back （2013年校赛 二分图+博弈）

Back

Time Limit: 2000 MS Memory Limit: 65536 K Total Submit: 7(2 users) Total Accepted: 1(1 users) Rating: Special Judge: No

Description

"Go" is a board game for two players that originated inChina more than 2,500 years ago. The game is noted for being rich in strategy despite its relatively simple rules. Xiaodao && xiaodai were playing "Go", but xiaodao always lose, so she came up with a brand new game on the board so called "Back". The stones are already on the board at the beginning. The two players alternately take back the stones. The first player can take back any stone on the board, and from the second turn, the player can only take back the stone which are 4-adjacent with the previous one. Xiaodao moves first. The player who can not take back any stone under this rule will be lose. Of couse, xiaodao want to predict weather she could win or not, no matter what decisions the opponent going to take. We assume both player will play optimally.

Input

The first line contains the number of test casesT(1 ≤ T ≤ 100), then each case contain one line two numbern, m — the size of the board(1 ≤ n, m ≤ 32). The nextn lines, each line contains m characters "."(empty) or "x"(stone) — the corresponding board at the beginning of the game.

Output

For each of the test cases numbered in order from 1 to T, output "Case #", followed by the case number, followed by ": " and a "Yes"(win) or "No"(lose).

Sample Input

3 1 1 . 3 3 .x. xxx .x. 4 5 .xx.. .x.xx .xx.x xxx.x

Sample Output

Case #1: No Case #2: Yes Case #3: No

Source

"科林明伦杯"哈尔滨理工大学第三届ACM程序设计团队赛

Author

xiaodao

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
#define clr(x) memset(x,0,sizeof(x))
struct node{
    int to,next;
}qq[50*50*50*50];
int head[50*50*50];
int tot;
void add(int s,int u)
{
    qq[tot].to=u;
    qq[tot].next=head[s];
    head[s]=tot++;
}
int l[50*50*50];
int v[50*50*50];
int find(int x){
    int k,i;
    for(i=head[x];i;i=qq[i].next){
        k=qq[i].to;
        if(!v[k]){
            v[k]=1;
            if(l[k]==0||find(l[k])){
                l[k]=x;
                return 1;
            }
        }
    }
    return 0;
}
char map[50][50];
int vis[50][50];
int num[50][50];
int p,q;
int ok(int x,int y){
    if(x>=0&&x<p&&y>=0&&y<q)return 1;
    return 0;
}
struct node1{
    int x,y;
}t,tt;
int main(){
    int T,i,j,ca=1;
    scanf("%d",&T);
    queue<struct node1>que;
    while(T--){
        int flag=0;
        scanf("%d%d",&p,&q);
        clr(map);
        clr(vis);
        for(i=0;i<p;i++){
            scanf("%s",&map[i]);
        }
        int tmp=1;
        for(i=0;i<p;i++){
            for(j=0;j<q;j++)
                num[i][j]=tmp++;
        }
        int fff=0;
        for(i=0;i<p;i++){
            for(j=0;j<q;j++){
                if(!que.empty())que.pop();
                if(map[i][j]=='x'&&!vis[i][j]){

                  fff++;
                    clr(head);
                    clr(l);
                    tot=1;
                    vis[i][j]=1;
                    t.x=i;t.y=j;
                    que.push(t);
                    int res=1;
                    while(!que.empty()){
                        tt=que.front();
                        que.pop();
                        if(ok(tt.x+1,tt.y)&&map[tt.x+1][tt.y]=='x'){
                            add(num[tt.x][tt.y],num[tt.x+1][tt.y]);
                            add(num[tt.x+1][tt.y],num[tt.x][tt.y]);
                            if(!vis[tt.x+1][tt.y]){
                                res++;vis[tt.x+1][tt.y]=1;t.x=tt.x+1;t.y=tt.y;que.push(t);
                            }
                        }
                        if(ok(tt.x-1,tt.y)&&map[tt.x-1][tt.y]=='x'){
                            add(num[tt.x-1][tt.y],num[tt.x][tt.y]);
                            add(num[tt.x][tt.y],num[tt.x-1][tt.y]);
                            if(!vis[tt.x-1][tt.y]){
                                res++;vis[tt.x-1][tt.y]=1;t.x=tt.x-1;t.y=tt.y;que.push(t);
                            }

                        }
                        if(ok(tt.x,tt.y+1)&&map[tt.x][tt.y+1]=='x'){
                            add(num[tt.x][tt.y],num[tt.x][tt.y+1]);
                            add(num[tt.x][tt.y+1],num[tt.x][tt.y]);
                            if(!vis[tt.x][tt.y+1]){
                                res++;vis[tt.x][tt.y+1]=1;t.x=tt.x;t.y=tt.y+1;que.push(t);
                            }
                        }
                        if(ok(tt.x,tt.y-1)&&map[tt.x][tt.y-1]=='x'){
                            add(num[tt.x][tt.y],num[tt.x][tt.y-1]);
                            add(num[tt.x][tt.y-1],num[tt.x][tt.y]);
                            if(!vis[tt.x][tt.y-1]){
                                 res++;vis[tt.x][tt.y-1]=1;t.x=tt.x;t.y=tt.y-1;que.push(t);
                            }
                        }

                    }
                     int sum=0;
                        int n=num[p-1][q-1];
                        for(int k=1;k<=n;k++){
                            clr(v);
                            if(find(k))sum++;
                        }

                        if(sum!=res){flag=1;goto loop;}
                }
            }
        }
    loop:    if(!flag||fff==0)printf("Case #%d: No\n",ca++);
            else printf("Case #%d: Yes\n",ca++);
    }
    return 0;
}