bzoj 1880 最短路

　　我们首先知道，答案肯定是最短路图中的某段公共链，那么设(x,y)为x到y的最短路，那么答案为((s1,t1)+(s2,t2)-min((s1,s2)+(t1,t2),(s1,t2),(s2,t1))) div 2，这两种情况分别可以画成下面的图

　　这样就可以十分明显的看出了。因为中间的一段算了两遍，所以要div 2。

/**************************************************************
    Problem:
    User: BLADEVIL
    Language: Pascal
    Result: Accepted
    Time: ms
    Memory: kb
****************************************************************/

//By BLADEVIL
var
    n, m                :longint;
    s1, t1, s2, t2      :longint;
    pre, other, len     :array[..] of longint;
    last                :array[..] of longint;
    flag                :array[..] of boolean;
    vis                 :array[..] of boolean;
    l                   :longint;
    dis, que            :array[..] of longint;
    d1, d2, d3, d4      :array[..] of longint;
    ans                 :longint;

function min(a,b:longint):longint;
begin
    if a>b then exit(b) else exit(a);
end;

function max(a,b:longint):longint;
begin
    if a>b then exit(a) else exit(b);
end;

procedure connect(x,y,z:longint);
begin
    inc(l);
    pre[l]:=last[x];
    last[x]:=l;
    other[l]:=y;
    len[l]:=z;
end;

procedure init;
var
    i                   :longint;
    x, y, z             :longint;
begin
    read(n,m,s1,t1,s2,t2);
    l:=;
    for i:= to m do
    begin
        read(x,y,z);
        connect(x,y,z);
        connect(y,x,z);
    end;
    fillchar(vis,sizeof(vis),true);
end;

procedure spfa(x:longint);
var
    q, p                :longint;
    h, t, cur           :longint;
begin
    filldword(dis,sizeof(dis) div ,maxlongint div );
    h:=; t:=;
    que[]:=x; dis[x]:=;
    while h<>t do
    begin
        h:=h mod +;
        cur:=que[h];
        flag[cur]:=false;
        q:=last[cur];
        while q<> do
        begin
            if vis[q] then
            begin
                p:=other[q];
                if dis[p]>dis[cur]+len[q] then
                begin
                    dis[p]:=dis[cur]+len[q];
                    if not flag[p] then
                    begin
                        t:=t mod +;
                        que[t]:=p;
                        flag[p]:=true;
                    end;
                end;
            end;
            q:=pre[q];
        end;
    end;
end;

procedure main;
var
    ans1, ans2          :longint;
    tot1, tot2          :longint;
    i                   :longint;
    q, p                :longint;

begin
    spfa(s1);
    ans1:=dis[t1];
    for i:= to n do d1[i]:=dis[i];

    spfa(s2);
    ans2:=dis[t2];
    for i:= to n do d2[i]:=dis[i];

    spfa(t1);
    for i:= to n do d3[i]:=dis[i];

    spfa(t2);
    for i:= to n do d4[i]:=dis[i];

    fillchar(vis,sizeof(vis),false);
    for i:= to n do
    begin
        q:=last[i];
        while q<> do
        begin
            p:=other[q];
            if d1[i]+len[q]+d3[p]=ans1 then
            begin
                vis[q]:=true;
                vis[q xor ]:=true;
            end;
            if d2[i]+len[q]+d4[p]=ans2 then
            begin
                vis[q]:=true;
                vis[q xor ]:=true;
            end;
            q:=pre[q];
        end;
    end;
    spfa(s1);
    tot1:=dis[s2];
    tot2:=dis[t2];
    spfa(t1);
    tot1:=tot1+dis[t2];
    tot2:=tot2+dis[s2];;
    ans:=(ans1+ans2-min(tot1,tot2)) div ;
    ans:=max(ans,);
    writeln(ans);
end;

begin
    init;
    main;
end.