链接：Link 耗时：0.028s

一道简单的动态规划，主要思路就是：用 $f[i,j]$ 表示到达 $(i,j)$ 的最长路径的长度。找到每个最高点，从其开始向四周的低处搜索。如果该点已搜过并且f值大于当前长度则退出回溯。直到达到某个最低点为止。

不多说了，直接上代码：

const    delta :array [1..4, 1..2] of integer = ((-1, 0), (1, 0), (0, 1), (0, -1)); //四个方向向量var    _: Integer;    name: string;    n, m, i, j, x: Integer;    ans: longint;    map: array [0..101, 0..101] of integer;    f: array [1..100, 1..100] of longint;function max(x, y: longint): longint; inline;begin    if x>y then exit(x) else exit(y);end;function can(x, y: integer): Boolean; inline; //判断是否是最高点var    i: Integer;    tx, ty: integer;begin    can := true;    for i := 1 to 4 do    begin        tx := x + delta[i, 1];        ty := y + delta[i, 2];        can := can and (map[x, y] >= map[tx, ty]);        if not can then break;    end;end;procedure dp(x, y: integer; len: longint); //回溯进行动态规划var    i: Integer;    tx, ty: integer;begin    inc(len);    if f[x, y] > len then exit;    f[x, y] := len;    ans := max(ans, len);    for i := 1 to 4 do    begin        tx := delta[i, 1] + x;        ty := delta[i, 2] + y;        if (tx = 0) or (tx > n) or (ty = 0) or (ty > m) then continue;        if map[x, y] <= map[tx, ty] then continue;        dp(tx, ty, len);    end;end;procedure ReadAndProcessName; //处理那行该死的名字！！var    s: string;    i: integer;begin    readln(s);    i := 1;    name := '';    n := 0;    m := 0;    while s[i] <> ' ' do    begin        name := name + s[i];        inc(i);    end;    inc(i);    while s[i] <> ' ' do    begin        n := n * 10 + ord(s[i]) - ord('0');        inc(i);    end;    inc(i);    while i <= length(s) do    begin        m := m * 10 + ord(s[i]) - ord('0');        inc(i);    end;end;begin    assign(input, 'main.in');reset(input);    assign(output, 'main.out');rewrite(output);    readln(_);    while _>0 do    begin        dec(_);        fillchar(map, sizeof(map), 0);        ReadAndProcessName;        for i := 1 to n do            for j := 1 to m do            begin                read(x);                map[i, j] := x+1;            end;        readln;        fillchar(f, sizeof(f), 0);        ans := 0;        for i := 1 to n do            for j := 1 to m do                if can(i, j) then                    dp(i, j, 0);        writeln(name, ': ', ans);    end;    close(input);close(output);end.