poj3169 [differential constraint + spfa]

Problem link http://poj.org/problem?id=3169

Topic meaning:

Some cows are arranged in a straight line by serial number.

There are two requirements, the distance between A and B should not exceed X, and the other is that the distance between C and D should not be less than Y, which is the maximum possible distance. If there is no output -1, if you can arrange the output -2 arbitrarily, otherwise output the largest distance.

First about differential constraints: https://blog.csdn.net/consciousman/article/details/53812818

After understanding the difference constraint, you can know the typical difference constraint + spfa of the question.

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 const int INF=0x3f3f3f3f;
 6 struct pot{
 7     int to;
 8     int next;
 9     int len;
10 }edge[20005];
11 int next[1005];
12 int d[1005];
13 int cnt[1005];
14 bool vis[1005];
15 int que[1005];
16 int tot=0,front=1,back=0;
17 int n,m,q;
18 void add(int x,int y,int t)
19 {
20     edge[tot].to=y;
21     edge[tot].len=t;
22     edge[tot].next=next[x];
23     next[x]=tot++;
24 }
25 bool spfa()
26 {
27     d[1]=0;
28     vis[1]=true;
29     que[front++]=1;
30     cnt[1]++;
31     while(front!=back)
32     {
33         back++;
34         if(back>=1005)back=0;
35         vis[que[back]]=false;
36         for(int i = next[que[back]];i!=-1;i=edge[i].next)
37         {
38             if(d[edge[i].to]>d[que[back]]+edge[i].len)
39             {
40                 d[edge[i].to]=d[que[back]]+edge[i].len;
41                 if(!vis[edge[i].to])
42                 {
43                     que[front++]=edge[i].to;
44                     if(front>=1005)front=0;
45                     vis[edge[i].to]=true;
46                     cnt[edge[i].to]++;
47                     if(cnt[edge[i].to]>n)return false;
48                 }
49             }
50         }
51     }
52     return true;
53 }
54 int main()
55 {
56     scanf("%d%d%d",&n,&m,&q);
57     for(int i = 1  ;i <= n ; i++)
58     {
59         next[i]=-1;
60         d[i]=INF;
61         vis[i]=false; 
62         cnt[i]=0;
63     }
64     for(int i = 0  ; i < m ; i++)
65     {
66         int a,b,c;
67         scanf("%d%d%d",&a,&b,&c);
68         if(a>b)
69         {
70             int t=a;
71             a=b;
72             b=t;
73         }
74         add(a,b,c);
75     }
76     for(int i = 0  ; i < q ; i++)
77     {
78         int a,b,c;
79         scanf("%d%d%d",&a,&b,&c);
80         if(b>a)
81         {
82             int t=a;
83             a=b;
84             b=t;
85         }
86         add(a,b,-c);
87     }
88     if(!spfa())printf("-1\n");
89     else if(d[n]==INF)printf("-2\n");
90     else printf("%d\n",d[n]);
91     return 0;
92 }

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