uva 11280(Dijkstra+递推)

题意：在一张有重边的有向图中有若干城市，有若干个询问每次告诉你最多能经过q个城市，让你求最多经过q个城市从1号城市到n号城市的最短路是什么？

思路：很容易想到用Dijkstra稍作变形就能解决这一题，只要在我们计算最短路的dis数组中再增加一维，表示最多经过的城市数，这样每次放进队列中的第二个状态变成两个，最后再遍历一下dis数组求一下最小值就能得出答案了。

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <cmath>
 5 #include <cstring>
 6 #include <algorithm>
 7 #include <queue>
 8 #include <stack>
 9 #include <vector>
10 #define MP(a, b) make_pair(a, b)
11 #define PB(a) push_back(a)
12 
13 using namespace std;
14 
15 typedef long long ll;
16 typedef pair<int ,int> pii;
17 typedef pair<unsigned int, unsigned int> puu;
18 typedef pair<int ,double> pid;
19 typedef pair<ll, int> pli;
20 
21 const int INF = 0x3f3f3f3f;
22 const double eps = 1e-6;
23 const int LEN = 110;
24 struct V{char name[22];};
25 typedef struct{int v, n;}St;
26 typedef pair<int, St> pis;
27 V vex[LEN];
28 int n, m, dis[LEN][LEN];
29 vector<pii> Map[LEN];
30 
31 struct cmp{
32     bool operator ()(pis a, pis b){return a.first>b.first;}
33 };
34 
35 void Dijkstra(int s)
36 {
37     priority_queue<pis, vector<pis>, cmp> q;
38     int vis[LEN][LEN] = {0};St temp;
39     memset(dis, 0x3f, sizeof dis);
40     dis[s][0] = 0;
41     temp.v = s;temp.n = 0;
42     q.push(MP(dis[s][0], temp));
43     while(!q.empty()){
44         pis nvex = q.top(); q.pop();
45         St nv = nvex.second;
46         if(vis[nv.v][nv.n])continue;
47         vis[nv.v][nv.n] = 1;
48         for(int i=0; i<Map[nv.v].size(); i++){
49             int x = Map[nv.v][i].first, y = Map[nv.v][i].second;
50             if(dis[x][nv.n+1]>dis[nv.v][nv.n]+y){
51                 dis[x][nv.n+1] = dis[nv.v][nv.n]+y;
52                 temp.v = x;temp.n = nv.n+1;
53                 q.push(MP(dis[x][nv.n+1], temp));
54             }
55         }
56     }
57 }
58 
59 int main()
60 {
61 //    freopen("in.txt", "r", stdin);
62 
63     int T, val, a, b, q, qn;
64     char from[22], to[22];
65     scanf("%d", &T);
66     for(int kase = 1; kase<=T; kase++){
67         for(int i=0; i<LEN; i++)Map[i].clear();
68         scanf("%d", &n);
69         for(int i=1; i<=n; i++){
70             scanf("%s", vex[i].name);
71         }
72         scanf("%d", &m);
73         for(int i=0; i<m; i++){
74             scanf("%s%s%d", from, to, &val);
75             for(int j=1; j<=n; j++){
76                 if(!strcmp(vex[j].name, from))a = j;
77                 if(!strcmp(vex[j].name, to))b = j;
78             }
79             Map[a].PB(MP(b, val));
80         }
81         Dijkstra(1);
82         printf("Scenario #%d\n", kase);
83         scanf("%d", &qn);
84         for(int i=0; i<qn; i++){
85             scanf("%d", &q);
86             int ans = INF;
87             for(int j=0; j<=q+1; j++){
88                 ans = min(ans, dis[n][j]);
89             }
90             if(ans!=INF)printf("Total cost of flight(s) is $%d\n", ans);
91             else printf("No satisfactory flights\n");
92         }
93         if(kase!=T)printf("\n");
94     }
95     return 0;
96 }
View Code

 

posted @ 2014-01-11 21:31  张小豪  阅读(424)  评论(0)  编辑  收藏  举报