HDOJ题目3440 House Man(差分约束)
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House Man
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2256 Accepted Submission(s): 896
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.
Each test case begins with a line containing two integers N (1 ≤ N ≤ 1000) and D (1 ≤ D ≤1000000). 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.
3 4 4 20 30 10 40 5 6 20 34 54 10 15 4 2 10 20 16 13
Case 1: 3 Case 2: 3 Case 3: -1
#include<stdio.h> #include<string.h> #include<queue> #include<iostream> #include<stdlib.h> #define INF 1<<30 #define min(a,b) (a>b?b:a) #define max(a,b) (a>b?a:b) using namespace std; struct s { int id,num; }b[1010]; int n,m,vis[1010],head[1010],cnt,dis[1010],out[1010]; int cmp(const void *a,const void *b) { return (*(struct s *)a).num-(*(struct s *)b).num; } struct node { int u,v,w,next; }edge[2020]; void add(int u,int v,int w) { edge[cnt].u=u; edge[cnt].v=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++; } int spfa(int s) { int i; for(i=1;i<=n;i++) dis[i]=INF; memset(vis,0,sizeof(vis)); vis[s]=1; dis[s]=0; queue<int>q; q.push(s); memset(out,0,sizeof(out)); while(!q.empty()) { int u=q.front(); q.pop(); vis[u]=0; out[u]++; if(out[u]>n) return 0; for(i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; if(dis[v]>dis[u]+edge[i].w) { dis[v]=dis[u]+edge[i].w; if(!vis[v]) { vis[v]=1; q.push(v); } } } } return 1; } int main() { int t,c=0; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); int i,j; memset(head,-1,sizeof(head)); cnt=0; for(i=1;i<=n;i++) { scanf("%d",&b[i].num); b[i].id=i; } qsort(b+1,n,sizeof(b[1]),cmp); for(i=2;i<=n;i++) { add(i,i-1,-1); if(b[i].id<b[i-1].id) { add(b[i].id,b[i-1].id,m); // add(i,i-1,-1); } else { add(b[i-1].id,b[i].id,m); // add(i-1,i,-1); } } // for(i=1;i<=n;i++) // { // add(n+1,i,0); // } int s=min(b[1].id,b[n].id); int t=max(b[1].id,b[n].id); int ans=spfa(s); printf("Case %d: ",++c); if(ans) { printf("%d\\n",dis[t]); } else printf("-1\\n"); } }
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