POJ2449 Remmarguts' Date

Posted 2020-10-05 嘒彼小星

Remmarguts‘ Date

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 31080		Accepted: 8486

Description

"Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly touching his little ducks‘ head, he told them a story.

"Prince Remmarguts lives in his kingdom UDF – United Delta of Freedom. One day their neighboring country sent them Princess Uyuw on a diplomatic mission."

"Erenow, the princess sent Remmarguts a letter, informing him that she would come to the hall and hold commercial talks with UDF if and only if the prince go and meet her via the K-th shortest path. (in fact, Uyuw does not want to come at all)"

Being interested in the trade development and such a lovely girl, Prince Remmarguts really became enamored. He needs you - the prime minister‘s help!

DETAILS: UDF‘s capital consists of N stations. The hall is numbered S, while the station numbered T denotes prince‘ current place. M muddy directed sideways connect some of the stations. Remmarguts‘ path to welcome the princess might include the same station twice or more than twice, even it is the station with number S or T. Different paths with same length will be considered disparate.

Input

The first line contains two integer numbers N and M (1 <= N <= 1000, 0 <= M <= 100000). Stations are numbered from 1 to N. Each of the following M lines contains three integer numbers A, B and T (1 <= A, B <= N, 1 <= T <= 100). It shows that there is a directed sideway from A-th station to B-th station with time T.

The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).

Output

A single line consisting of a single integer number: the length (time required) to welcome Princess Uyuw using the K-th shortest path. If K-th shortest path does not exist, you should output "-1" (without quotes) instead.

Sample Input

2 2
1 2 5
2 1 4
1 2 2

Sample Output

14

Source

POJ Monthly,Zeyuan Zhu

 

【题解】

这个破题数据是肯定有错误的

s与t相等时，最短路是0，但数据并没有算上

调了一晚上，早上看了网上博客才发现的这个问题

 

A*算法，估价函数是当前路径长度+这个点到终点的最短路长度。

每次取值最小的进行拓展，每拓展到终点一次代表找到下一条最短路。

第k次拓展到终点表示找到第k短路

 

  1 #include <iostream>
  2 #include <cstdlib>
  3 #include <cstring>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <queue>
  7 #include <vector>
  8 
  9 inline void read(int &x)
 10 {
 11     x = 0;char ch = getchar(), c = ch;
 12     while(ch < ‘0‘ || ch > ‘9‘)c = ch, ch = getchar();
 13     while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘, ch = getchar();
 14     if(c == ‘-‘)x = -x;
 15 } 
 16 
 17 const int MAXN = 50000 + 10;
 18 const int MAXM = 500000 + 10;
 19 
 20 struct Edge
 21 {
 22     int u,v,w,next;
 23     Edge(int _u, int _v, int _w, int _next){u = _u;v = _v; w = _w; next = _next;}
 24     Edge(){}
 25 }edge[MAXM << 2], edge2[MAXM << 2];
 26 int head[MAXN], head2[MAXN], cnt, cnt2;
 27 
 28 inline void insert(int a, int b, int c)
 29 {
 30     edge[++cnt] = Edge(a,b,c,head[a]);
 31     head[a] = cnt;
 32 }
 33 
 34 inline void insert2(int a, int b, int c)
 35 {
 36     edge2[++cnt2] = Edge(a,b,c,head2[a]);
 37     head2[a] = cnt2;
 38 }
 39 
 40 int n,m,s,t,k;
 41 
 42 int b[MAXN], d[MAXN];
 43 
 44 struct Node
 45 {
 46     int p, d;
 47     Node(int _p, int _d){p = _p;d = _d;}
 48     Node(){}    
 49 };
 50 
 51 struct cmp
 52 {
 53     bool operator()(Node a, Node b)
 54     {
 55         return a.d > b.d;
 56     }    
 57 };
 58 
 59 std::priority_queue<Node, std::vector<Node>, cmp> q;
 60 
 61 void dijkstra()
 62 {
 63     while(q.size())q.pop(); 
 64     memset(d, 0x3f, sizeof(d));
 65     q.push(Node(t, 0));
 66     d[t] = 0;
 67     register Node tmp;
 68     while(q.size())
 69     {
 70         tmp = q.top(), q.pop();
 71         if(b[tmp.p])continue;
 72         b[tmp.p] = 1;
 73         for(register int pos = head[tmp.p];pos;pos = edge[pos].next)
 74         {
 75             int v = edge[pos].v;
 76             if(b[v])continue;
 77             if(d[v] > tmp.d + edge[pos].w)
 78                 d[v] = tmp.d + edge[pos].w, q.push(Node(v, d[v]));
 79         }
 80     }
 81 }
 82 
 83 struct cmpp
 84 {
 85     bool operator()(Node a, Node b)
 86     {
 87         return a.d + d[a.p] > b.d + d[b.p];
 88     }    
 89 };
 90 
 91 std::priority_queue <Node, std::vector<Node>, cmpp> q2;
 92 
 93 int step;//拓展次数 
 94 
 95 void abfs()
 96 {
 97      step = 0;
 98      while(q2.size())q2.pop(); 
 99      q2.push(Node(s, 0));
100      register Node tmp;
101      if(d[s] == d[0])
102      {
103          printf("-1\n");
104          return;
105      }
106      while(q2.size())
107      {
108          tmp = q2.top(),q2.pop();
109          if(tmp.p == t) ++ step;
110          if(step == k)
111          {
112              printf("%d\n", tmp.d);
113              return;
114          }
115          for(register int pos = head2[tmp.p];pos;pos = edge2[pos].next)
116          {
117              int v = edge2[pos].v;
118              q2.push(Node(edge2[pos].v, tmp.d + edge2[pos].w));
119          }
120      }
121      printf("-1\n");
122 }
123 
124 int main()
125 {
126     register int tmp1, tmp2, tmp3;
127     while(scanf("%d %d", &n, &m) != EOF && (n + m))
128     {
129         memset(b, 0, sizeof(b));
130         memset(head, 0, sizeof(head));
131         memset(head2, 0, sizeof(head2));
132         memset(edge, 0, sizeof(edge));
133         memset(edge2, 0, sizeof(edge2));
134         cnt = cnt2 = 0;
135         for(register int i = 1;i <= m;++ i)
136         {
137             read(tmp1), read(tmp2), read(tmp3);
138             insert2(tmp1, tmp2, tmp3);
139             insert(tmp2, tmp1, tmp3);
140         }
141         read(s), read(t), read(k);
142         if(t == s)++ k;
143         dijkstra();
144         abfs();
145     }
146     return 0;
147 } 

POJ2249