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.