POJ3621-Sightseeing Cows

二分密度+bellman-ford判断是否存在负权回路。

a,b,e分别代表每条边的起点、终点、权值，mid为当前密度，cost为点的快乐值

在二分密度时，要注意更新边权为e[i]=e[i]*mid-cost[b[i]]

 

[pascal 代码]

VAR
        A,B,COST:ARRAY[1..100000]OF LONGINT;
        DIS,E,EE:ARRAY[1..100000]OF DOUBLE;
        N,M:LONGINT;
PROCEDURE INIT;
 VAR
        I:LONGINT;
 BEGIN
        READLN(N,M);
        FOR I:=1 TO N DO READLN(COST[I]);
        FOR I:=1 TO M DO READLN(A[I],B[I],E[I]);
        EE:=E;
 END;
FUNCTION BELLMAN_FORD:BOOLEAN;
 VAR
        I,J:LONGINT;
        FLAG:BOOLEAN;
 BEGIN
        FILLCHAR(DIS,SIZEOF(DIS),0);
        FOR I:=1 TO N DO
                BEGIN
                        FLAG:=FALSE;
                        FOR J:=1 TO M DO IF DIS[A[J]]+E[J]<DIS[B[J]] THEN
                                BEGIN
                                        DIS[B[J]]:=DIS[A[J]]+E[J];
                                        FLAG:=TRUE;
                                END;
                        IF NOT FLAG THEN EXIT(TRUE);
                END;
        EXIT(FALSE);
 END;
PROCEDURE MAIN;
 VAR
        L,R,MID:DOUBLE;
        I:LONGINT;
 BEGIN
        L:=0;R:=1997;
        WHILE L+1E-6<R DO
                BEGIN
                        MID:=(L+R)/2;
                        FOR I:=1 TO M DO E[I]:=EE[I]*MID-COST[B[I]];
                        IF BELLMAN_FORD THEN R:=MID ELSE L:=MID;
                END;
        WRITELN(R:0:2);
 END;
BEGIN
        INIT;
        MAIN;
END.