Hdu 6071 Lazy Running【同余最短路】

Lazy Running

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 927    Accepted Submission(s): 381

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 thanK 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 thanK.

 

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

12000 600 650 535 380

 

Sample Output

2165
Hint
The best path is 2-1-4-3-2.

题目大意：

现在有4个点，我们一开始在点2处 ，我们希望最终也回到点2处 ，问我们怎样走，使得总路径和大于等于K并且最小。

思路：

我们知道，我们可以选以2为起点，1作为另一个点（或者是3的话也是相同的道理），使得我们的主人公一直徘徊在这条路径上，那么我们最终的Ans，可以表示为2*W(1,2)+Len这里Len的长度明显在区间：【0,2*W(1,2)】之间，那么我们就去跑一个同余的最短路即可。

那么我们SPFA跑最短路，处理出数组Dist【i】【Len】表示，从2出发，到点i处，长度%2*W(1,2)为Len的最短路。

那么对于结果，我们枚举区间【0,2*W(1,2)】的值作为Len，然后向上去添加 2*W(1,2)即可。

对于Dist【2】【i】，我们考虑：

如果Dist【2】【i】>=K,那么更新ans=min（ans，Dist【2】【i】）；

否则我们设定Ned=K-Dist【2】【i】，表示缺少的路径长度，我们向上去添加2*W(1,2)即可，添加的长度为：Ned/2*W(1,2)*(2*W(1,2))+Ned%(2*W(1,2))==0?0:2*W(1,2)。

其内容不难理解。

注意初始化数组和INF要足够大，就没有别的什么坑点了。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<queue>using namespace std;#define ll __int64const ll INF=2e18;struct node{    int u;    ll len;} now,nex;ll mod;ll mp[5][5];ll vis[5][70050];ll dist[5][70050];void SPFA(){    queue<node>s;    memset(vis,0,sizeof(vis));    for(int j=1; j<=4; j++)    {        for(int k=0; k<=70000; k++)        {            dist[j][k]=INF;        }    }    dist[2][0]=0;    now.u=2,now.len=0;    s.push(now);    while(!s.empty())    {        now=s.front();        // vis[now.u][now.len%mod]=0;        s.pop();        for(int v=1; v<=4; v++)        {            if(mp[now.u][v]!=INF)            {                nex.u=v;                nex.len=now.len+mp[now.u][v];                if(dist[v][nex.len%mod]>now.len+mp[now.u][v])                {                    dist[v][nex.len%mod]=now.len+mp[now.u][v];                    if(vis[v][nex.len%mod]==0)                    {                        vis[v][nex.len%mod]=1;                        s.push(nex);                    }                }            }        }    }}int main(){    int t;    scanf("%d",&t);    while(t--)    {        ll K,a,b,c,d;        scanf("%I64d%I64d%I64d%I64d%I64d",&K,&a,&b,&c,&d);        for(int i=1; i<=4; i++)        {            for(int j=1; j<=4; j++)            {                mp[i][j]=INF;            }        }        mp[1][2]=mp[2][1]=a;        mp[2][3]=mp[3][2]=b;        mp[3][4]=mp[4][3]=c;        mp[4][1]=mp[1][4]=d;        mod=2*mp[1][2];        SPFA();        ll ans=INF;        for(int i = 0; i < mod; i++)        {            if(dist[2][i]>=K)            {                ans=min(ans,dist[2][i]);            }            else            {                ll ned=K-dist[2][i];                ll tmp;                if(ned%mod==0)tmp=0;                else tmp=mod;                ans=min(ans,dist[2][i]+ned/mod*mod+tmp);            }        }        printf("%I64d\n",ans);    }}