HDU 6071 同余最短路 spfa

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

). You can run more distance between two adjacent checkpoints, but only the distance saved at the system will be counted.

HDU 6071 同余最短路 spfa

Input

The first line of the input contains an integer

, 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 }