（最短路）AtCoder Beginner Contest 061 D - Score Attack

D - Score Attack

Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

There is a directed graph with N vertices and M edges. The i-th edge (1≤i≤M) points from vertex ai to vertex bi, and has a weight ci. We will play the following single-player game using this graph and a piece.

Initially, the piece is placed at vertex 1, and the score of the player is set to 0. The player can move the piece as follows:

  • When the piece is placed at vertex ai, move the piece along the i-th edge to vertex bi. After this move, the score of the player is increased by ci.

The player can end the game only when the piece is placed at vertex N. The given graph guarantees that it is possible to traverse from vertex 1 to vertex N.

When the player acts optimally to maximize the score at the end of the game, what will the score be? If it is possible to increase the score indefinitely, print inf.

Constraints

  • 2≤N≤1000
  • 1≤M≤min(N(N−1),2000)
  • 1≤ai,bi≤N(1≤i≤M)
  • ai≠bi(1≤i≤M)
  • ai≠aj or bi≠bj(1≤i<j≤M)
  • −109≤ci≤109(1≤i≤M)
  • ci is an integer.
  • In the given graph, there exists a path from vertex 1 to vertex N.

Input

Input is given from Standard Input in the following format:

N M  
a1 b1 c1  
a2 b2 c2
:  
aM bM cM

Output

Print the maximum possible score at the end of the game, if it is finite. If it is possible to increase the score indefinitely, print inf.

Sample Input 1

Copy
3 3
1 2 4
2 3 3
1 3 5

Sample Output 1

Copy
7

There are two ways to move the piece to vertex N=3:

  • vertex 1 → vertex 2 → vertex 3 : score 4+3=7
  • vertex 1 → vertex 3 : score 5

Thus, the maximum possible score at the end of the game is 7.

Sample Input 2

Copy
2 2
1 2 1
2 1 1

Sample Output 2

Copy
inf

It is possible to increase the score indefinitely by alternating between vertex 1 and 2.

Sample Input 3

Copy
6 5
1 2 -1000000000
2 3 -1000000000
3 4 -1000000000
4 5 -1000000000
5 6 -1000000000

Sample Output 3

Copy
-5000000000

 

由于出现负权值边，需要使用Bellman-Ford算法判断是否有和为正的环，并且此环必须还在某条能到n点的路径上。

其中，为了方便处理，将边的权值先都乘-1，以将最长路变成最短路问题。

利用Bellman-Ford算法循环顶点数-1之后如果还会更新dis，就说明有负环的性质，再加上延展这样的环的范围，如果能延展到顶点n，就为inf的情况，不然就输出最初记录的答案。

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <queue>
 8 #include <set>
 9 #include <map>
10 #include <list>
11 #include <stack>
12 #include <vector>
13 #define mp make_pair
14 typedef long long ll;
15 typedef unsigned long long ull;
16 const int MAX=2e3+5;
17 const int NAX=1e3+5;
18 const ll INF=1e18+5;
19 const double M=4e18;
20 using namespace std;
21 const int MOD=1e9+7;
22 typedef pair<int,int> pii;
23 const double eps=0.000000001;
24 int n,m;
25 ll dis[NAX];
26 int from[MAX],to[MAX];
27 bool neg[MAX];
28 ll cost[MAX];
29 ll an;
30 int main()
31 {
32     scanf("%d%d",&n,&m);
33     for(int i=1;i<=m;++i)
34     {
35         scanf("%d%d%lld",&from[i],&to[i],&cost[i]);
36         cost[i]=-cost[i];
37     }
38     fill(dis+1,dis+1+n,INF);
39     dis[1]=0;
40     for(int i=0;i<n-1;++i)
41     {
42         for(int j=1;j<=m;++j)
43         {
44             if(dis[from[j]]==INF)
45                 continue;
46             if(dis[to[j]]>dis[from[j]]+cost[j])
47             {
48                 dis[to[j]]=dis[from[j]]+cost[j];
49             }
50         }
51     }
52     an=dis[n];
53     for(int i=0;i<n-1;++i)
54     {
55         for(int j=1;j<=m;++j)
56         {
57             if(dis[from[j]]==INF)
58                 continue;
59             if(dis[to[j]]>dis[from[j]]+cost[j])
60             {
61                 neg[to[j]]=true;
62                 dis[to[j]]=dis[from[j]]+cost[j];
63             }
64             if(neg[from[j]])
65                 neg[to[j]]=true;
66         }
67     }
68     if(neg[n])
69         printf("inf\n");
70     else
71         printf("%lld\n",-an);
72 }
View Code

 

posted @ 2017-05-13 23:12  perplex  阅读(338)  评论(0)  编辑  收藏  举报