HDU 6071 Lazy Running （同余最短路 dij）

最新推荐文章于 2019-08-18 14:36:24 发布

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最新推荐文章于 2019-08-18 14:36:24 发布

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文章标签： java

原文链接：http://www.cnblogs.com/jianrenfang/p/7599448.html

本文介绍了一道关于寻找满足特定条件的最短路径问题——LazyRunning，并提供了详细的算法解析及实现代码。该问题涉及无限循环路径上的最短距离求解，采用同余最短路径方法解决。

Lazy Running

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1384 Accepted Submission(s): 597

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

Input

The first line of the input contains an integer

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】四个点连成环，相邻两个点之间有距离，问从点 1 出发回到点1 ，总距离超过K 的最短路是多少。

【分析】像这种无限走下去的题，可以用同余最短路来解。点1相邻的两条边，设最短的那条长度为，m,那么存在一条长度为x的回到1节点的路，就一定存在长度为x+2*m的路。

dis[i][j]表示到达i点总长度%2*m==j的最短路，然后dij就行了。

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 6e4+50;;
const int M = 255;
const int mod = 19260817;
const int mo=123;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
typedef pair<ll,int>P;
int n,s;
ll dis[6][N];
ll k,edg[6][6],m,ans;
void dij(int s){
    priority_queue<P,vector<P>,greater<P> >q;
    for(int i=0;i<4;i++){
        for(int j=0;j<=m;j++){
            dis[i][j]=1e18;
        }
    }
    q.push(P(0LL,s));
    while(!q.empty()){
        ll w=q.top().first;
        int u=q.top().second;
        q.pop();
        if(u==s){
            if(w<k){
                ans=min(ans,w+((k-w-1)/m+1)*m);
            }
            else ans=min(ans,w);
        }
        for(int i=0;i<4;i++){
            if(!edg[u][i])continue;
            ll d=w+edg[u][i];
            if(dis[i][d%m]>d){
                dis[i][d%m]=d;
                q.push(P(d,i));
            }
        }
    }
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        ans=1e18;
        scanf("%lld",&k);
        for(int i=0;i<4;i++){
            scanf("%lld",&edg[i][(i+1)%4]);
            edg[(i+1)%4][i]=edg[i][(i+1)%4];
        }
        m=2*min(edg[1][0],edg[1][2]);
        ans=((k-1)/m+1)*m;
        dij(1);
        printf("%lld\n",ans);
    }
    return 0;
}