P1073-最优贸易

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define _for(i,a,b) for(int i = (a);i < b;i ++)
 4 #define _rep(i,a,b) for(int i = (a);i > b;i --)
 5 #define INF 0x3f3f3f3f
 6 #define pb push_back
 7 #define maxn 5053900
 8 typedef long long ll;
 9 
10 inline ll read()
11 {
12     ll ans = 0;
13     char ch = getchar(), last = ‘ ‘;
14     while(!isdigit(ch)) last = ch, ch = getchar();
15     while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - ‘0‘, ch = getchar();
16     if(last == ‘-‘) ans = -ans;
17     return ans;
18 }
19 inline void write(ll x)
20 {
21     if(x < 0) x = -x, putchar(‘-‘);
22     if(x >= 10) write(x / 10);
23     putchar(x % 10 + ‘0‘);
24 }
25 int tot,ver[maxn],Next[maxn],head[maxn],val[maxn];
26 void add(int x,int y,int t)
27 {
28     ver[++tot] = y,Next[tot] = head[x],head[x] = tot,val[tot] = t;
29 }
30 int n,m;
31 int v[maxn];
32 typedef pair<int,int> P;
33 int d[maxn];
34 void shortest_path(int s)
35 {
36     priority_queue<P> que;
37     memset(d,-0x3f,sizeof(d));
38     d[s] = 0;
39     que.push(P {0,s});
40 
41     while(!que.empty())
42     {
43         P p = que.top();
44         que.pop();
45         int v = p.second;
46         if(d[v] > p.first) continue;
47         for(int i = head[v]; i; i = Next[i])
48         {
49             int y = ver[i];
50             int w = val[i];
51             if(d[y] < d[v] + w)
52             {
53                 d[y] = d[v] + w;
54                 que.push(P {d[y],y});
55             }
56         }
57     }
58 }
59 int main()
60 {
61      n = read();
62      m = read();
63      _for(i,1,n+1)
64          v[i] = read();
65      _for(i,1,m+1)
66      {
67         int a = read();
68         int b = read();
69         int c = read(); 
70         if(c==1)
71         {
72             add(a,b,0);
73             add(n+a,n+b,0);
74             add(2*n+a,2*n+b,0);
75         }
76         else
77         {
78             add(a,b,0);
79             add(b,a,0);
80             add(n+a,n+b,0);
81             add(n+b,n+a,0);
82             add(2*n+b,2*n+a,0);
83             add(2*n+a,2*n+b,0);
84         }    
85     }
86     _for(i,1,n+1)
87      {
88          add(i,i+n,-v[i]);
89          add(i+n,i+n*2,v[i]);
90     }
91     shortest_path(1);
92     printf("%d\n",max(0,d[3*n]));
93     return 0;
94 }