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  • POJ 1125

    POJ1125   

    题目大意:在一幅有向图中,求在某个点出发,到达最远的点最少的时间是多少?

    解:暴力最短路即可…….

    View Code
      1 const
    2 maxn=100;
    3 bilibili=maxlongint >> 1;
    4 type
    5 data=record
    6 dest, next, cost: longint;
    7 end;
    8 var
    9 edge: array[1..maxn*maxn]of data;
    10 dist, vect, heap, poss: array[1..maxn]of longint;
    11 hptot, tot, n, who, ans: longint;
    12 procedure print;
    13 begin
    14 if ans=bilibili then
    15 writeln('disjoint')
    16 else writeln(who, ' ', ans);
    17 end;
    18
    19 procedure add(x, y, z: longint);
    20 begin
    21 inc(tot);
    22 with edge[tot] do
    23 begin
    24 dest := y;
    25 cost := z;
    26 next := vect[x];
    27 vect[x] := tot;
    28 end;
    29 end;
    30
    31 procedure init;
    32 var
    33 tmp, i, j, x, y: longint;
    34 begin
    35 fillchar(vect, sizeof(vect), 0);
    36 tot := 0;
    37 ans := bilibili;
    38 readln(n);
    39 if n=0 then
    40 begin
    41 close(input); close(output);
    42 halt;
    43 end;
    44 for i := 1 to n do
    45 begin
    46 read(tmp);
    47 for j := 1 to tmp do
    48 begin
    49 read(x, y);
    50 add(i, x, y);
    51 end;
    52 readln;
    53 end;
    54 end;
    55
    56 procedure up(x: longint);
    57 var
    58 i, tmp: longint;
    59 begin
    60 i := x;
    61 tmp := heap[x];
    62 while i>1 do
    63 begin
    64 if dist[tmp]<dist[heap[i >> 1]] then
    65 begin
    66 heap[i] := heap[i >> 1];
    67 poss[heap[i]] := i;
    68 i := i >> 1;
    69 end
    70 else break;
    71 end;
    72 heap[i] := tmp;
    73 poss[tmp] := i;
    74 end;
    75
    76 procedure down(x: longint);
    77 var
    78 i, j, tmp: longint;
    79 begin
    80 i := x;
    81 tmp := heap[x];
    82 while i << 1 <= hptot do
    83 begin
    84 j := i << 1;
    85 if (j+1<=hptot)and(dist[heap[j]] > dist[heap[j+1]]) then inc(j);
    86 if dist[tmp]>dist[heap[j]] then
    87 begin
    88 heap[i] := heap[j];
    89 poss[heap[i]] := i;
    90 i := j;
    91 end
    92 else break;
    93 end;
    94 heap[i] := tmp;
    95 poss[tmp] := i;
    96 end;
    97
    98 procedure dijkstra(b: longint);
    99 var
    100 i, j, u: longint;
    101 begin
    102 fillchar(poss, sizeof(poss), 0);
    103 filldword(dist, sizeof(dist)>>2, bilibili);
    104 hptot := 1;
    105 dist[b] := 0;
    106 heap[1] := b;
    107 poss[b] := 1;
    108 repeat
    109 u := heap[1];
    110 heap[1] := heap[hptot];
    111 poss[heap[1]] := 1;
    112 dec(hptot);
    113 down(1);
    114 i := vect[u];
    115 while i<>0 do
    116 with edge[i] do
    117 begin
    118 if dist[u] + cost < dist[dest] then
    119 begin
    120 dist[dest] := dist[u] + cost;
    121 if poss[dest]=0 then
    122 begin
    123 inc(hptot);
    124 heap[hptot] := dest;
    125 poss[dest] := hptot;
    126 up(hptot);
    127 end
    128 else
    129 up(poss[dest]);
    130 end;
    131 i := next;
    132 end;
    133 until hptot<1;
    134 end;
    135
    136 procedure main;
    137 var
    138 i, j, tmp: longint;
    139 begin
    140 while true do
    141 begin
    142 init;
    143 for i := 1 to n do
    144 begin
    145 tmp := 0;
    146 dijkstra(i);
    147 for j := 1 to n do
    148 begin
    149 if tmp<dist[j] then tmp := dist[j];
    150 if tmp=bilibili then break;
    151 end;
    152 if ans>tmp then
    153 begin
    154 ans := tmp;
    155 who := i;
    156 end;
    157 end;
    158 print;
    159 end;
    160 end;
    161
    162 begin
    163 //assign(input,'aaa.in'); reset(input);
    164 main;
    165 end.



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  • 原文地址:https://www.cnblogs.com/wmzisfoolish/p/2435155.html
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