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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#425550#4773. Piece it togetherGe_QingHuiAC ✓138ms26652kbC++142.8kb2024-05-30 13:32:232024-05-30 13:32:24

Judging History

你现在查看的是最新测评结果

  • [2024-05-30 13:32:24]
  • 评测
  • 测评结果:AC
  • 用时:138ms
  • 内存:26652kb
  • [2024-05-30 13:32:23]
  • 提交

answer

#pragma GCC optmize(2,3,"Ofast","inline","unroint-loops")
#include<bits/stdc++.h>
using namespace std;
const int N=5e5+9;
const int M=5e6+9;
int n,m,cnt,mat[N],id[2][509][509],s,t;
const int INF=0x3f3f3f3f;
bool vst[N];
char mp[509][509];
struct Dinic{
    struct Edge{int u,v,nxt;} Eg[M];
    int hd[N],tot=1;
    void add(int x,int y,int z){
	    tot++; Eg[tot]=(Edge){y,z,hd[x]}; hd[x]=tot;
        tot++; Eg[tot]=(Edge){x,0,hd[y]}; hd[y]=tot;
    }
    void clear(){
        tot=1;
        for(int i=1;i<=cnt;i++) hd[i]=0;
    }
    int d[N];
    queue<int> q;
    bool bfs(){
	    for(int i=1;i<=cnt;i++) d[i]=0;
	    while(!q.empty()) q.pop();
	    q.push(s); d[s]=1;
	    while(!q.empty()){
		    int pre=q.front(); q.pop();
		    for(int i=hd[pre],u;i;i=Eg[i].nxt)
		        if(Eg[i].v && !d[u=Eg[i].u]){
		    	    q.push(u); d[u]=d[pre]+1;
		    	    if(u==t) return true;
			    }
	    }
	return false;
    }
    int dfs(int x,int flow){
	    if(x==t) return flow;
	    int rest=flow;
	    for(int i=hd[x],u;i && rest;i=Eg[i].nxt)
	        if(Eg[i].v && d[u=Eg[i].u]==d[x]+1 && d[x]!=0){
	    	    int k=dfs(u,min(rest,Eg[i].v));
	    	    if(!k) d[u]=-INF;
			    Eg[i].v-=k;
			    Eg[i^1].v+=k;
			    rest-=k;
		    }
	    return flow-rest;
    }
    int dinic(){
        int ans=0;
	    while(bfs()){
		    int cnt=0;
		    while(1){
			    cnt=dfs(s,INF);
			    if(cnt==0) break;
		        ans+=cnt;
		    }
	    }
        return ans;
    }
} T;
const int dx[5]={0,0,1,-1};
const int dy[5]={1,-1,0,0};
void solve(){
    cnt=0; cin>>n>>m;
    
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++) cin>>mp[i][j];
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            if(mp[i][j]=='.') continue;
            if(mp[i][j]=='W') id[0][i][j]=++cnt;
            if(mp[i][j]=='B') id[0][i][j]=++cnt,id[1][i][j]=++cnt;
        }
    s=++cnt; t=++cnt;
    T.clear();
    int sB=0,sW=0;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            if(mp[i][j]=='B'){
                for(int k=0;k<4;k++){
                    int ni=i+dx[k],nj=j+dy[k];
                    if(1<=ni && ni<=n && 1<=nj && nj<=m && mp[ni][nj]=='W'){
                        T.add(id[(k<2)][i][j],id[0][ni][nj],1);
                    }
                }
                sB++;
                T.add(s,id[0][i][j],1);
                T.add(s,id[1][i][j],1);
            }
            else if(mp[i][j]=='W'){
                sW++;
                T.add(id[0][i][j],t,1);
            }
        }
    if(sB*2!=sW) puts("NO");
    else if(T.dinic()!=sW) puts("NO");
    else puts("YES");
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);

    int T; cin>>T;
    while(T--) solve();
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 3660kb

input:

2
3 4
BWW.
WWBW
..WB
3 3
W..
BW.
WBW

output:

YES
NO

result:

ok 2 token(s): yes count is 1, no count is 1

Test #2:

score: 0
Accepted
time: 138ms
memory: 26652kb

input:

70
3 4
BWW.
WWBW
..WB
3 3
W..
BW.
WBW
1 1
B
3 3
...
.W.
...
2 2
W.
BW
2 3
.W.
WBW
1 3
WBW
2 5
.W.W.
WB.BW
2 2
WW
.B
2 2
WB
..
3 3
WWB
BWW
WWB
3 5
.W.WB
WBW.W
...WB
4 5
..W..
.WBW.
WBWBW
.WBW.
3 9
BWW...W..
WWBW..BW.
..WB..WBW
4 12
BWWBWWBWWBWW
WWBWWBWWBWWB
BWWBWWBWWBWW
WWBWWBWWBWWB
7 7
BWWBBWW
WBWWW...

output:

YES
NO
NO
NO
YES
NO
NO
YES
NO
NO
NO
NO
YES
NO
YES
YES
NO
YES
YES
YES
NO
YES
YES
NO
NO
NO
NO
NO
NO
NO
NO
NO
YES
YES
YES
NO
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
NO
YES
YES
YES
NO
YES
NO
YES
NO
NO
NO
NO
NO
YES
NO
YES
YES
NO
NO
NO
NO
NO

result:

ok 70 token(s): yes count is 34, no count is 36