BZOJ2763 [JLOI2011] 飞行路线

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2763

Description

Alice和Bob现在要乘飞机旅行，他们选择了一家相对便宜的航空公司。该航空公司一共在n个城市设有业务，设这些城市分别标记为0到n-1，一共有m种航线，每种航线连接两个城市，并且航线有一定的价格。Alice和Bob现在要从一个城市沿着航线到达另一个城市，途中可以进行转机。航空公司对他们这次旅行也推出优惠，他们可以免费在最多k种航线上搭乘飞机。那么Alice和Bob这次出行最少花费多少？

Input

数据的第一行有三个整数，n,m,k，分别表示城市数，航线数和免费乘坐次数。
第二行有两个整数，s,t，分别表示他们出行的起点城市编号和终点城市编号。(0<=s,t<n)
接下来有m行，每行三个整数，a,b,c，表示存在一种航线，能从城市a到达城市b，或从城市b到达城市a，价格为c。(0<=a,b<n,a与b不相等，0<=c<=1000)

Output

只有一行，包含一个整数，为最少花费。
 

距离增加一维，d[i][j]表示第i个点距起点，免费j次的最短距离

Dijkstra 224ms，SPFA6500ms……差点卡进Status第一页，据说SPFA用SLF会更快

第一次使用指针保存邻接表

另外Status第二页怎么都是一堆熟悉的人啊……还有一堆高一好可怕……耀良，泽川，PoPoQQQ神犇（Orz），SA（Orz），KPM（Orz）

Dijkstra，224ms

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <queue>
 6 #define rep(i,l,r) for(int i=l; i<=r; i++)
 7 #define clr(x,y) memset(x,y,sizeof(x))
 8 #define travel(x) for(Edge *i=last[x]; i; i=i->pre)
 9 using namespace std;
10 const int INF = 0x3f3f3f3f;
11 const int maxn = 10010;
12 struct Edge{
13     Edge *pre; int to,cost;
14 }edge[100010];
15 int n,m,k,s,t,x,y,z,ans,d[maxn][11];
16 Edge *pt,*last[maxn];
17 struct node{
18     int x,k,d;
19     node(){}
20     node(int _x,int _k,int _d) : x(_x), k(_k), d(_d){};
21     inline bool operator < (const node &_Tp) const{
22         return d > _Tp.d;
23     }
24 }now;
25 priority_queue <node> q;
26 inline int read(){
27     int ans = 0, f = 1;
28     char c = getchar();
29     while (!isdigit(c)){
30         if (c == '-') f = -1;
31         c = getchar();
32     }
33     while (isdigit(c)){
34         ans = ans * 10 + c - '0';
35         c = getchar();
36     }
37     return ans * f;
38 }
39 inline void addedge(int x,int y,int z){
40     pt->pre = last[x];
41     pt->to = y;
42     pt->cost = z;
43     last[x] = pt++;
44 }
45 void dijkstra(){
46     clr(d,INF); clr(d[s],0);
47     q.push(node(s,0,0));
48     rep(i,1,k) q.push(node(s,i,0));
49     while (!q.empty()){
50         now = q.top(); q.pop();
51         if (now.d != d[now.x][now.k]) continue;
52         travel(now.x){
53             if (d[now.x][now.k] + i->cost < d[i->to][now.k]){
54                 d[i->to][now.k] = d[now.x][now.k] + i->cost;
55                 q.push(node(i->to,now.k,d[i->to][now.k]));
56             }
57             if (d[now.x][now.k] < d[i->to][now.k+1] && now.k < k){
58                 d[i->to][now.k+1] = d[now.x][now.k];
59                 q.push(node(i->to,now.k+1,d[i->to][now.k+1]));
60             }
61         }
62     }
63 }
64 int main(){
65     n = read(); m = read(); k = read(); s = read(); t = read();
66     clr(last,0); pt = edge;
67     rep(i,1,m){
68         x = read(); y = read(); z = read();
69         addedge(x,y,z); addedge(y,x,z);
70     }
71     dijkstra();
72     ans = d[t][0];
73     rep(i,1,k) ans = min(ans,d[t][i]);
74     printf("%d\n",ans);
75     return 0;
76 }
View Code

SPFA，6500ms

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <queue>
 6 #define rep(i,l,r) for(int i=l; i<=r; i++)
 7 #define clr(x,y) memset(x,y,sizeof(x))
 8 #define travel(x) for(Edge *i=last[x]; i; i=i->pre)
 9 using namespace std;
10 const int INF = 0x3f3f3f3f;
11 const int maxn = 10010;
12 struct Edge{
13     Edge *pre; int to,cost;
14 }edge[100010];
15 int n,m,k,s,t,x,y,z,ans,d[maxn][11];
16 bool isin[maxn][11];
17 Edge *pt,*last[maxn];
18 struct node{
19     int x,k;
20     node(){}
21     node(int _x,int _k) : x(_x), k(_k){};
22 }now;
23 queue <node> q;
24 inline int read(){
25     int ans = 0, f = 1;
26     char c = getchar();
27     while (!isdigit(c)){
28         if (c == '-') f = -1;
29         c = getchar();
30     }
31     while (isdigit(c)){
32         ans = ans * 10 + c - '0';
33         c = getchar();
34     }
35     return ans * f;
36 }
37 inline void addedge(int x,int y,int z){
38     pt->pre = last[x];
39     pt->to = y;
40     pt->cost = z;
41     last[x] = pt++;
42 }
43 void spfa(){
44     clr(d,INF); clr(d[s],0); clr(isin,0);
45     q.push(node(s,0)); isin[s][0] = 1;
46     rep(i,1,k) q.push(node(s,i)), isin[s][i] = 1;
47     while (!q.empty()){
48         now = q.front(); q.pop(); isin[now.x][now.k] = 0;
49         travel(now.x){
50             if (d[now.x][now.k] + i->cost < d[i->to][now.k]){
51                 d[i->to][now.k] = d[now.x][now.k] + i->cost;
52                 if (!isin[i->to][now.k]){
53                     isin[i->to][now.k] = 1;
54                     q.push(node(i->to,now.k));
55                 }
56             }
57             if (d[now.x][now.k] < d[i->to][now.k+1] && now.k < k){
58                 d[i->to][now.k+1] = d[now.x][now.k];
59                 if (!isin[i->to][now.k+1]){
60                     isin[i->to][now.k+1] = 1;
61                     q.push(node(i->to,now.k+1));
62                 }
63             }
64         }
65     }
66 }
67 int main(){
68     n = read(); m = read(); k = read(); s = read(); t = read();
69     clr(last,0); pt = edge;
70     rep(i,1,m){
71         x = read(); y = read(); z = read();
72         addedge(x,y,z); addedge(y,x,z);
73     }
74     spfa();
75     ans = d[t][0];
76     rep(i,1,k) ans = min(ans,d[t][i]);
77     printf("%d\n",ans);
78     return 0;
79 }
View Code

 

posted on 2015-12-10 21:25  ACMICPC  阅读(193)  评论(0)  编辑  收藏  举报

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