UESTC Training for Graph Theory——L、House Man

House Man

Time Limit: 1000 ms Memory Limit: 65536 kB Solved: 43 Tried: 595

Description

In somewhere, there is a crazy super man. He can’t fly, but he could jump from housetop to housetop. Today he plans to use N houses to hone his house hopping skills. He will start at the shortest house and make N-1 jumps, with each jump taking him to a taller house than the one he is jumping from. When finished, he will have been on every house exactly once, traversing them in increasing order of height, and ending up on the tallest house. 
The man can travel for at most a certain horizontal distance D in a single jump. To make this as much fun as possible, the crazy man want to maximize the distance between the positions of the shortest house and the tallest house. 
The crazy super man have an ability—move houses. So he is going to move the houses subject to the following constraints:
1. All houses are to be moved along a one-dimensional path. 
2. Houses must be moved at integer locations along the path, with no two houses at the same location. 
3. Houses must be arranged so their moved ordering from left to right is the same as their ordering in the input. They must NOT be sorted by height, or reordered in any way. They must be kept in their stated order. 
4. The super man can only jump so far, so every house must be moved close enough to the next taller house. Specifically, they must be no further than D apart on the ground (the difference in their heights doesn't matter). 
Given N houses, in a specified order, each with a distinct integer height, help the super man figure out the maximum possible distance they can put between the shortest house and the tallest house, and be able to use the houses for training. 

Input

In the first line there is an integer T, indicates the number of test cases.(T<=50)
Each test case begins with a line containing two integers N (1 ≤ N ≤ 10000) and D (1 ≤ D ≤100000). The next line contains N integer, giving the heights of the N houses, in the order that they should be moved. Within a test case, all heights will be unique. 

Output

For each test case , output “Case #d: “first where d is the case number counted from one, then output a single integer representing the maximum distance between the shortest and tallest house, subject to the constraints above, or -1 if it is impossible to lay out the houses. Do not print any blank lines between answers.

Sample Input

3
4 4 
20 30 10 40 
5 6 
20 34 54 10 15 
4 2 
10 20 16 13 

Sample Output

Case #1: 3
Case #2: 3
Case #3: -1

Source

2010 ACM-ICPC MU Training - Host by FZU

 

/*算法思想：  题意复杂，就不简述了  查分约束，建图是这样的：  1、每相邻的两个 house 的距离差不小于 1  2、第 i 大的 house 和第 i+1 大的 house 的距离不大于D  这样，我们建好图之后对 house 最小的点跑一遍最短路，记录下它到 house 最大的点的最短路，就是最终的答  案了。  由于边少点多，所以一定是用spfa跑最短路  这道题最麻烦的是判断无解的情况，刚开始用的每个点入队超过 n 次就无解（出现了负环），但是超时是肯定的  毕竟我们OJ的上的数据强的离谱。然后听老师们讲了dfs判重，还是不怎么搞懂，后来才知道原来spfa可以用栈来  实现，这样就方便判重了，遇到遍历过的点肯定是存在负环了，那么一定无解，不过貌似在没有负环的时候这样的  spfa跑的时间比用队列跑的spfa长*/#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define INF 0x7fffffff#define N 40000using namespace std;struct data1{    int st,en,val,next;} edge[N];struct data2{    int id,height;} house[N];int n,d,tot,head[N],min_id,max_id,dis[N];;bool vs[N];void add_edge(int st,int en,int val)  //添加边{    edge[tot].st=st;    edge[tot].en=en;    edge[tot].val=val;    edge[tot].next=head[st];    head[st]=tot++;}bool cmp(data2 a,data2 b){    return a.height<b.height;}bool dfs_spfa(int v)  //用栈的dfs_spfa跑最短路{    if(vs[v]) return false;    vs[v]=true;    for(int i=head[v];i!=-1;i=edge[i].next)        if(dis[edge[i].en]>dis[v]+edge[i].val)        {            dis[edge[i].en]=dis[v]+edge[i].val;            if(!dfs_spfa(edge[i].en)) return false;        }    vs[v]=false;    return true;}int main(){    int t;    scanf("%d",&t);    for(int ca=1;ca<=t;ca++)    {        memset(head,-1,sizeof(head));        tot=1;        scanf("%d%d",&n,&d);        for(int i=1;i<=n;i++)        {            scanf("%d",&house[i].height);            house[i].id=i;        }        sort(house+1,house+n+1,cmp);        for(int i=2;i<=n;i++)        {            int v1=min(house[i-1].id,house[i].id);            int v2=max(house[i-1].id,house[i].id);            add_edge(v1,v2,d);            add_edge(i,i-1,-1);        }        printf("Case #%d: ",ca);        min_id=min(house[1].id,house[n].id);        max_id=max(house[1].id,house[n].id);        memset(vs,0,sizeof(vs));        for(int i=1;i<=n;i++)            dis[i]=INF;        dis[min_id]=0;        if(!dfs_spfa(min_id))            printf("-1\n");        else            printf("%d\n",dis[max_id]);    }    return 0;}