function lca(a,b:longint):longint;
var
        i,j:longint;
begin
        if deep[a]<depp[b] then swap(a,b);
        i:=0;
        while (1<<i)<=deep[a] do inc(i);
        dec(i);
        for j:=i downto 0 do
        if deep[a]-(1<<j)>=deep[b] then a:=f[a,j];
        if a=b then exit
 
(a);
        for j:=i downto 0 do
        if (f[a,j]<>-1)and(f[a,j]<>f[b,j]) then
        begin
                a:=f[a,j];
                b:=f[b,j];
        end;
        exit(f[a,0]);
end;

tarjan求lca

procedure tarjan(x:longint);
var
        i,v:longint;
begin
        vis[x]:=1;
        fa[x]:=x;
        i:=head1[x];
        while
 
 i<>0 do//第一個鄰接表儲存詢問
        begin
                v:=go1[i];
                if vis[v] then
                ans[id[i]]:=find(v);
                i:=next1[i];
        end;
        i:=head[x];//第二個鄰接表儲存圖
        while i<>0 do
        begin
                v:=go[i];
                if not vis[v] then
 

                begin
                        dis[v]:=dis[x]+len[i];
                        tarjan(v);
                        fa[v]:=x;
                end;
                i:=next[i];
        end;
end;

3.最小（大）生成樹
kruskal

var f:array[1..1001] of longint;
    a:array[0..500001,1..3] of longint;
    i,j,k,n,m,x,y,c:longint;
    s:int64;
procedure qsort(l,r:longint);
    var x,y,m:longint;
    begin
        x:=l;
        y:=r;
        m:=a[(l+r) div 2,3];
        repeat
            while a[x,3]<m do inc(x);
            while a[y,3]>m do dec(y);
            if x<=y then
            begin
                a[0]:=a[x];
                a[x]:=a[y];
                a[y]:=a[0];
                inc(x);dec(y);
            end;
        until x>y;
        if l<y then qsort(l,y);
        if r>x then qsort(x,r);
    end;
function fu(z:longint):longint;
    var x,y:longint;
    begin
        y:=z;
        while y<>f[y] do y:=f[y];

        while z<>y do
        begin
            x:=f[z];
            f[z]:=y;
            z:=x;
        end;
    exit(y);
    end;
begin
    readln(n,m);
    for i:=1 to m do
    begin
        readln(x,y,c);
        a[i,1]:=x;
        a[i,2]:=y;
        a[i,3]:=c;
    end;
    qsort(1,m);

    for i:=1 to n do f[i]:=i;

    s:=0;
    for i:=1 to m do
    begin
        j:=fu(a[i,1]);
        k:=fu(a[i,2]);
        if j<>k then
        begin
            s:=s+a[i,3];
            f[j]:=k;
        end;
    end;
    writeln(s);
end.

比較簡單，常用
4.二分圖匹配（匈牙利）（判斷二分圖在最上面的那個網址，當然自己隨手打個染色也是可以的。）

bool dfs(int u){
    for(int i = head[u]; i != 0; i = e[i].nex){
        int v = e[i].v;
        if(!vis[v]){
            vis[v] = true;
            if(link[v] == -1 || dfs(link[v])){
                link[v] = u;
                return true;
            }
        }
    }
    return false;
}

5.最短路
自行百度。（我覺得不會考dij+heap）