• POJ 3159


    POJ 3159

    题目大意:给n个小朋友们分糖果,其中有m个关系a b c,代表A允许B比自己最多多C个糖果,求第一个小朋友到第n个小朋友的糖果数的差最大是多少(满足这m个条件下)

    解:题目有隐藏题意,第n个小朋友(即班长要比1多),所以这些条件就是一个差分约束,然后求1到n的最短路即可(dist[x]表示1允许x比自己多多少,把关系传递,显然是一个最短路模型,如果存在一条1到x的路比已知短,显然要更新才能满足题意。),初始化全部为bilibili,dist[1]为0,鬼畜的是即使是spfa+slf+queue也会tle,但是用spfa+stack就会很快(我刷上第四名),300+ms,dij+heap也要500ms。

    spfa+stack
     1 //poj 3159
    2 const
    3 maxn=31111;
    4 maxm=151111;
    5 bilibili=maxlongint >> 1;
    6 type
    7 data=record
    8 cost, dest, next: longint;
    9 end;
    10 var
    11 edge: array[1..maxm]of data;
    12 q, vect, dist: array[1..maxn]of longint;
    13 visit: array[1..maxn]of boolean;
    14 n, m, tot, ans: longint;
    15 procedure add(x, y, z: longint);
    16 begin
    17 inc(tot);
    18 with edge[tot] do begin
    19 dest := y;
    20 cost := z;
    21 next := vect[x];
    22 vect[x] := tot;
    23 end;
    24 end;
    25
    26 procedure init;
    27 var
    28 i, x, y, z: longint;
    29 begin
    30 tot := 0; ans := 0;
    31 fillchar(vect, sizeof(vect), 0);
    32 readln(n, m);
    33 for i := 1 to m do begin
    34 readln(x, y, z);
    35 add(x, y, z);
    36 end;
    37 end;
    38
    39 function spfa(source: longint): longint;
    40 var
    41 i, u, head, tail, key: longint;
    42 begin
    43 fillchar(dist, sizeof(dist), 0);
    44 fillchar(visit, sizeof(visit), 0);
    45 tail := 1;
    46 q[tail] := source; visit[source] := true;
    47 filldword(dist, sizeof(dist)>>2, bilibili);
    48 dist[source] := 0;
    49 repeat
    50 u := q[tail];
    51 dec(tail);
    52 visit[u] := false;
    53 i := vect[u];
    54 while i<>0 do
    55 with edge[i] do begin
    56 if dist[dest]>dist[u]+cost then begin
    57 dist[dest] := dist[u] + cost;
    58 if not visit[dest] then begin
    59 visit[dest] := true;
    60 inc(tail);
    61 q[tail] := dest;
    62 end;
    63 end;
    64 i := next;
    65 end;
    66 until tail<=0;
    67 spfa := dist[n];
    68 end;
    69
    70 procedure main;
    71 begin
    72 ans := spfa(1);
    73 end;
    74
    75 procedure print;
    76 begin
    77 writeln(ans);
    78 end;
    79
    80 begin
    81 assign(input,'1.txt'); reset(input);
    82 init;
    83 main;
    84 print;
    85 end.
    dijkstra+heap
      1 //poj 3159
    2 const
    3 maxn=31111;
    4 maxm=151111;
    5 bilibili=maxlongint >> 1;
    6 type
    7 data=record
    8 cost, dest, next: longint;
    9 end;
    10 var
    11 edge: array[1..maxm]of data;
    12 heap, poss, vect, dist: array[1..maxn]of longint;
    13 visit: array[1..maxn]of boolean;
    14 hptot, n, m, tot, ans: longint;
    15 procedure add(x, y, z: longint);
    16 begin
    17 inc(tot);
    18 with edge[tot] do begin
    19 dest := y;
    20 cost := z;
    21 next := vect[x];
    22 vect[x] := tot;
    23 end;
    24 end;
    25
    26 procedure init;
    27 var
    28 i, x, y, z: longint;
    29 begin
    30 tot := 0; ans := 0;
    31 fillchar(vect, sizeof(vect), 0);
    32 readln(n, m);
    33 for i := 1 to m do begin
    34 readln(x, y, z);
    35 add(x, y, z);
    36 end;
    37 end;
    38
    39 procedure up(x: longint);
    40 var
    41 i, tmp: longint;
    42 begin
    43 i := x;
    44 tmp := heap[x];
    45 while i>1 do begin
    46 if dist[tmp]<dist[heap[i >> 1]] then begin
    47 heap[i] := heap[i >> 1];
    48 poss[heap[i]] := i;
    49 i := i >> 1;
    50 end
    51 else break;
    52 end;
    53 heap[i] := tmp;
    54 poss[tmp] := i;
    55 end;
    56
    57 procedure down(X: longint);
    58 var
    59 i, j, tmp: longint;
    60 begin
    61 i := x;
    62 tmp := heap[x];
    63 while i << 1 <= hptot do begin
    64 j := i << 1;
    65 if (j+1<=hptot)and(dist[heap[j]]>dist[heap[j+1]]) then inc(j);
    66 if dist[tmp]>dist[heap[j]] then begin
    67 heap[i] := heap[j];
    68 poss[heap[i]] := i;
    69 i := j;
    70 end
    71 else break;
    72 end;
    73 heap[i] := tmp;
    74 poss[tmp] := i;
    75 end;
    76
    77 function dijkstra(source: longint): longint;
    78 var
    79 i, u: longint;
    80 begin
    81 fillchar(poss, sizeof(poss), 0);
    82 hptot := 0; poss[source] := -1; u := 1;
    83 filldword(dist, sizeof(dist)>>2, bilibili);
    84 dist[source] := 0;
    85 repeat
    86 if u=n then break;
    87 i := vect[u];
    88 while i<>0 do
    89 with edge[i] do begin
    90 if dist[dest]>dist[u]+cost then begin
    91 dist[dest] := dist[u] + cost;
    92 if poss[dest] = 0 then begin
    93 inc(hptot);
    94 heap[hptot] := dest; poss[dest] := hptot;
    95 up(hptot);
    96 end
    97 else up(poss[dest]);
    98 end;
    99 i := next;
    100 end;
    101 u := heap[1];
    102 poss[u] := -1;
    103 heap[1] := heap[hptot];
    104 poss[heap[1]] := 1;
    105 dec(hptot);
    106 down(1);
    107 until false;
    108 dijkstra := dist[n]-dist[source];
    109 end;
    110
    111 procedure main;
    112 begin
    113 ans := dijkstra(1);
    114 end;
    115
    116 procedure print;
    117 begin
    118 writeln(ans);
    119 end;
    120
    121 begin
    122 assign(input,'1.txt'); reset(input);
    123 init;
    124 main;
    125 print;
    126 end.




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  • 原文地址:https://www.cnblogs.com/wmzisfoolish/p/2435196.html
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