[二分答案] POJ2110 Mountain Walking

题目大意：

给你一个n×n的矩阵，让你找一条从(1,1)到(n,n)的一条路径，保证路径上经过点的最大值与最小值之差最小，输出这个差值。

题目分析：

典型的二分答案，枚举上下界区间，然后二分差值，求差值的最小值。

[pascal 代码]

CONST
        DX:ARRAY[1..4]OF -1..1=(0,0,1,-1);
        DY:ARRAY[1..4]OF -1..1=(-1,1,0,0);
TYPE
        LZY=RECORD
                X,Y:LONGINT;
        END;
VAR
        MAP:ARRAY[0..111,0..111]OF LONGINT;
        Q:ARRAY[0..100000]OF LZY;
        VIS:ARRAY[0..111,0..111]OF BOOLEAN;
        DOWN,UP,I,N,HEAD,TAIL,MIN,MAX,MINL:LONGINT;
PROCEDURE INIT;
 VAR
        I,J:LONGINT;
 BEGIN
        READLN(N);
        MAX:=0;
        MIN:=MAXLONGINT;
        FOR I:=1 TO N DO
                BEGIN
                        FOR J:=1 TO N DO
                                BEGIN
                                        READ(MAP[I,J]);
                                        IF MAP[I,J]<MIN THEN MIN:=MAP[I,J];
                                        IF MAP[I,J]>MAX THEN MAX:=MAP[I,J];
                                END;
                        READLN;
                END;
 END;
PROCEDURE INITQUE;
 BEGIN
        Q[1].X:=1;
        Q[1].Y:=1;
        VIS[1,1]:=TRUE;
        HEAD:=0;TAIL:=1;
 END;
FUNCTION BFS(DOWN,UP:LONGINT):BOOLEAN;
 VAR
        I,J:LONGINT;
        PRE,TMP:LZY;
 BEGIN
        FILLCHAR(VIS,SIZEOF(VIS),FALSE);
        INITQUE;
        IF MAP[1,1]>UP THEN EXIT(FALSE);
        IF MAP[1,1]<DOWN THEN EXIT(FALSE);
        WHILE HEAD<TAIL DO
                BEGIN
                        INC(HEAD);
                        PRE:=Q[HEAD];
                        FOR I:=1 TO 4 DO
                                BEGIN
                                        TMP.X:=PRE.X+DX[I];
                                        TMP.Y:=PRE.Y+DY[I];
                                        IF (TMP.X<1) OR (TMP.X>N) OR (TMP.Y<1) OR (TMP.Y>N) THEN CONTINUE;
                                        IF (MAP[TMP.X,TMP.Y]>=DOWN)AND(MAP[TMP.X,TMP.Y]<=UP)AND(NOT VIS[TMP.X,TMP.Y]) THEN
                                                BEGIN
                                                        INC(TAIL);
                                                        Q[TAIL]:=TMP;
                                                        VIS[TMP.X,TMP.Y]:=TRUE;
                                                        IF VIS[N,N] THEN EXIT(TRUE);
                                                END;
                                END;
                END;
        EXIT(FALSE);
 END;
PROCEDURE MAIN;
 VAR
        L,R,MID,I:LONGINT;
 BEGIN
        MINL:=MAXLONGINT;
        FOR I:=MIN TO MAX DO
        BEGIN
        L:=0;R:=MAX-MIN;
        WHILE L<R DO
                BEGIN
                        MID:=(L+R)>>1;
                        IF BFS(I,I+MID) THEN R:=MID ELSE L:=MID+1;
                END;
        IF L<MINL THEN MINL:=L;
        END;
        WRITELN(MINL);
 END;
BEGIN
        INIT;
        MAIN;
END.