[SGU]121. Bridges painting

Analysis

    刚拿到这道题的时候也想不出什么算法，然后去搜了一下才发现原来是简单的DFS构造……从任意奇数点开始DFS，对遍历到的边交替染色，如果这么做无解的话那么就No solution，否则输出解。证明似乎不那么容易，不过自己在纸上画一画应当能感觉出来吧……

Accepted Code

var    map,color:array[0..110,0..110] of longint;    deg:array[0..110] of longint;    n,i,j,t:longint;procedure dfs(u,c:longint);var    i,j:longint;begin    c:=3-c;    for i:=1 to deg[u] do    begin        j:=map[u,i];        if color[u,j]=0 then        begin            color[u,j]:=c;            color[j,u]:=c;            dfs(j,c);            c:=3-c;        end;    end;end;begin    readln(n);    fillchar(map,sizeof(map),0);    fillchar(deg,sizeof(deg),0);    fillchar(color,sizeof(color),0);    for i:=1 to n do    begin        read(t);        while t<>0 do        begin            inc(deg[i]);            map[i,deg[i]]:=t;            read(t);        end;    end;    for i:=1 to n do        if odd(deg[i]) then            dfs(i,1);    for i:=1 to n do        dfs(i,1);    for i:=1 to n do        if deg[i]>1 then        begin            t:=0;            for j:=1 to deg[i] do                t:=t or color[i,map[i,j]];            if t<>3 then            begin                writeln('No solution');                exit;            end;        end;    for i:=1 to n do    begin        for j:=1 to deg[i] do            write(color[i,map[i,j]],' ');        writeln(0);    end;end.