[SDOI2010]大陆争霸

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<vector>
 8 #include<queue>
 9 #include<stack>
10 #include<cctype>
11 using namespace std;
12 #define enter puts("")
13 #define space putchar(‘ ‘)
14 #define Mem(a) memset(a, 0, sizeof(a))
15 typedef long long ll;
16 typedef double db;
17 const int INF = 0x3f3f3f3f;
18 const db eps = 1e-8;
19 const int maxn = 3e3 + 5;
20 inline ll read()
21 {
22     ll ans = 0;
23     char ch = getchar(), last = ‘ ‘;
24     while(!isdigit(ch)) last = ch, ch = getchar();
25     while(isdigit(ch)) ans = (ans << 3) + (ans << 1) + ch - ‘0‘, ch = getchar();
26     if(last == ‘-‘) ans = -ans;
27     return ans;
28 }
29 inline void write(ll x)
30 { 
31     if(x < 0) putchar(‘-‘), x = -x;
32     if(x >= 10) write(x / 10);
33     putchar(x % 10 + ‘0‘);
34 }
35 
36 int n, m;
37 vector<int> v[maxn],c[maxn], su[maxn];        //su[i]代表有哪些城市控制城市i 
38 int cnt[maxn];        //记录入度 
39 
40 #define pr pair<int, int>
41 #define mp make_pair 
42 int d1[maxn], d2[maxn];
43 bool in[maxn];
44 void dijkstra(int s)
45 {
46     for(int i = 1; i <= n; ++i) d1[i] = INF;
47     d1[s] = 0;
48     priority_queue<pr, vector<pr>, greater<pr> > q;
49     q.push(mp(d1[s], s));
50     while(!q.empty())
51     {
52         pr node = q.top(); q.pop();
53         int now = node.second, dis = node.first;
54         if(in[now]) continue;
55         in[now] = 1;
56         for(int i = 0; i < (int)v[now].size(); ++i)
57         {
58             if(dis + c[now][i] < d1[v[now][i]])
59             {
60                 d1[v[now][i]] = dis + c[now][i];
61                 if(!cnt[v[now][i]]) q.push(mp(max(d1[v[now][i]], d2[v[now][i]]), v[now][i]));
62                 //只有入度为0的节点才可以被加入队列 
63             }
64         }
65         for(int i = 0; i < (int)su[now].size(); ++i)
66         {
67             cnt[su[now][i]]--;
68 //            d2[su[now][i]] = max(d2[su[now][i]], dis);
69             if(!cnt[su[now][i]])
70             {
71                 d2[su[now][i]] = dis;    
72 //                这句话和68句话是等价的，因为最后一个到达控制它的城市时间一定是最长的 
73                 q.push(mp(max(d1[su[now][i]], d2[su[now][i]]), su[now][i]));
74             }
75         }
76     }
77     write(max(d1[n], d2[n]));
78 }
79 
80 int main()
81 {
82     n = read(); m = read();
83     for(int i = 1; i <= m; ++i)
84     {
85         int x = read(), y = read(), w = read();
86         v[x].push_back(y); c[x].push_back(w);
87     }
88     for(int i = 1; i <= n; ++i)
89     {
90         cnt[i] = read();
91         for(int j = 1; j <= cnt[i]; ++j) su[read()].push_back(i);
92     }
93     dijkstra(1);
94     return 0;
95 }