Lazy Running

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 .

1
2000 600 650 535 380

2165


Hint

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

 

K是你要跑的最少长度，跑步必须跑完一段路，比如从1到2点，不能中途返回

要你求出大于等于K的最小值。

思路：与2这个点连接的有d1和d2这两条边，取小的那条 的两倍作为 m，你只要跑回2点之后，不够的话再跑m一直跑到够就ok了，但是在补m之前的跑路也要是最优的

用一个dis[i][j]表示到i点，且路程为j的时候距离，可以写成dis[i][j % m]，因为你乱跑跑到 i 用了 j 的距离，和你用最短路到达 i ，然后补m,补成 j 这两种走法是一样的，

这样mod m 不但可以省空间，还可以降低查询时间。

最后取查询一遍dis[1][j] (j 从 0 到 m)数组，取最小值，如果刚好是k就输出，如果不够，就补m。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<vector>
#include<queue>
using namespace std;
#define INF 2000000000000000000
#define ll long long
#define MAXN 100005
struct Edge{
int y;
ll w;
};
vector<Edge>e[5];
ll dis[4][MAXN],k,m;
struct par{
ll x;
int y;
bool operator < (const par &a) const {
return x>a.x;//最小值优先
}
};
inline void Dijkstra(int id){
priority_queue<par> q;
for(int i = 0;i < 4;i++){
for(int j = 0;j <= m;j++){
dis[i][j] = INF;
}
}
par p;
p.x = 0ll;
p.y = id;
q.push(p);
while(!q.empty()){
ll w = q.top().x;
int y = q.top().y;
q.pop();
if(w > dis[y][w % m])
// 如果这段路程已经比最优情况长了，那就不用继续下去了
continue;
for(int i = 0;i < e[y].size();i++){
int nxt = e[y][i].y;
ll dist = w + e[y][i].w;
if(dis[nxt][dist % m] > dist){
//由于最后都用 m 去补，所以dist可以直接mod m
dis[nxt][dist % m] = dist;	// 多的部分你换成去跑 m 和你真实去跑是一样的
par temp;
temp.x = dist;
temp.y = nxt;
q.push(temp);
}
}
}
}
int main(){
int t;
scanf("%d",&t);
while(t--){
int d1,d2,d3,d4;
memset(e,0,sizeof(e));
scanf("%lld %d %d %d %d",&k,&d1,&d2,&d3,&d4);
m = 2 * min(d1,d2);
e[0].push_back(Edge{1,d1});
e[1].push_back(Edge{0,d1});
e[1].push_back(Edge{2,d2});
e[2].push_back(Edge{1,d2});
e[2].push_back(Edge{3,d3});
e[3].push_back(Edge{2,d3});
e[0].push_back(Edge{3,d4});
e[3].push_back(Edge{0,d4});
Dijkstra(1);
ll ans = INF;
for(int i = 0;i < m;i++){
ll dist = k - dis[1][i];
if(dist <= 0){
ans = min(ans,dis[1][i]);
//如果刚好到起点跑完，那就是那么长，不用补 m
}else{
ans = min(ans,dis[1][i] + dist / m * m + (dist % m > 0) * m);
}
}
printf("%lldn",ans);
}
return 0;
}

最后

以上就是长情方盒为你收集整理的HDU 6071 最短路思维题 Lazy Running的全部内容，希望文章能够帮你解决HDU 6071 最短路思维题 Lazy Running所遇到的程序开发问题。

如果觉得靠谱客网站的内容还不错，欢迎将靠谱客网站推荐给程序员好友。