HDU 6071 同余最短路 spfa

Posted 2020-10-01 半根毛线code

Lazy Running

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 657    Accepted Submission(s): 284

Problem Description

In HDU, you have to run along the campus for 24 times, or you will fail in PE. According to the rule, you must keep your speed, and your running distance should not be less than K meters.

There are 4 checkpoints in the campus, indexed as p1,p2,p3 and p4. Every time you pass a checkpoint, you should swipe your card, then the distance between this checkpoint and the last checkpoint you passed will be added to your total distance.

The system regards these 4 checkpoints as a circle. When you are at checkpoint pi, you can just run to pi−1 or pi+1(p1 is also next to p4). You can run more distance between two adjacent checkpoints, but only the distance saved at the system will be counted.




Checkpoint p2 is the nearest to the dormitory, Little Q always starts and ends running at this checkpoint. Please write a program to help Little Q find the shortest path whose total distance is not less than K.

 

 

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there are 5 integers K,d1,2,d2,3,d3,4,d4,1(1≤K≤1018,1≤d≤30000), denoting the required distance and the distance between every two adjacent checkpoints.

 

 

Output

For each test case, print a single line containing an integer, denoting the minimum distance.

 

 

Sample Input

1 2000 600 650 535 380

 

 

Sample Output

2165

Hint

The best path is 2-1-4-3-2.

 

 

Source

2017 Multi-University Training Contest - Team 4

题意：求从2点回到2点的路径距离和大于等与K的最小值

题解：叉姐的分析

 1 #pragma comment(linker, "/STACK:102400000,102400000")
 2 #include <bits/stdc++.h>
 3 #include <cstdlib>
 4 #include <cstdio>
 5 #include <iostream>
 6 #include <cstdlib>
 7 #include <cstring>
 8 #include <algorithm>
 9 #include <cmath>
10 #include <cctype>
11 #include <map>
12 #include <set>
13 #include <queue>
14 #include <bitset>
15 #include <string>
16 #include <complex>
17 #define LL long long
18 #define mod 1000000007
19 using namespace std;
20 LL x,d12,d23,d34,d41;
21 int t;
22 struct node{
23     LL st,we,dis;
24 }exm,ok;
25 int vis[5][60004];
26 int dp[5][60004];
27 LL d[5][5];
28 LL bew;
29 queue<node> q;
30 void dij()
31 {
32     while(!q.empty())
33         q.pop();
34     exm.st=2;
35     exm.we=0;
36     exm.dis=0;
37     q.push(exm);
38     dp[2][0]=0;
39     vis[exm.st][exm.we]=1;
40     while(!q.empty()){
41         exm=q.front();
42         q.pop();
43         vis[exm.st][exm.we]=0;
44         LL now,ww;
45         now=(exm.st)%4+1;
46         ww=(exm.dis+d[exm.st][now])%(2*bew);
47         if(dp[now][ww]==-1||dp[now][ww]>=exm.dis+d[exm.st][now]){
48             dp[now][ww]=exm.dis+d[exm.st][now];
49             if(vis[now][ww]==0){
50             vis[now][ww]=1;
51             ok.st=now;
52             ok.we=ww;
53             ok.dis=exm.dis+d[exm.st][now];
54             q.push(ok);
55             }
56         }
57         now=(exm.st-1);
58         if(now==0)
59             now=4;
60         ww=(exm.dis+d[exm.st][now])%(2*bew);
61         if(dp[now][ww]==-1||dp[now][ww]>=exm.dis+d[exm.st][now]){
62             dp[now][ww]=exm.dis+d[exm.st][now];
63             if(vis[now][ww]==0){
64             vis[now][ww]=1;
65             ok.st=now;
66             ok.we=ww;
67             ok.dis=exm.dis+d[exm.st][now];
68             q.push(ok);
69         }
70         }
71     }
72 
73 }
74 int main()
75 {
76     scanf("%d",&t);
77     for(int i=1;i<=t;i++){
78         memset(vis,0,sizeof(vis));
79         memset(dp,-1,sizeof(dp));
80         scanf("%lld %lld %lld %lld %lld",&x,&d[1][2],&d[2][3],&d[3][4],&d[4][1]);
81         d[2][1]=d[1][2];
82         d[3][2]=d[2][3];
83         d[4][3]=d[3][4];
84         d[1][4]=d[4][1];
85         bew=min(d[2][1],d[2][3]);
86         dij();
87         LL ans=1e18+1;
88         for(LL j=0;j<2*bew;j++){
89             if(dp[2][j]==-1)
90                 continue;
91             LL zhong=(max(0LL,x-dp[2][j]))/(2*bew);
92           if(dp[2][j]+zhong*2*bew<x)
93             zhong++;
94          ans=min(ans,dp[2][j]+zhong*2*bew);
95        }
96        printf("%lld\n",ans);
97     }
98     return 0;
99 }