hdu 2363(枚举+最短路好题)
Cycling
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1247 Accepted Submission(s): 411
want to cycle to a programming contest. The shortest route to the
contest might be over the tops of some mountains and through some
valleys. From past experience you know that you perform badly in
programming contests after experiencing large differences in altitude.
Therefore you decide to take the route that minimizes the altitude
difference, where the altitude difference of a route is the difference
between the maximum and the minimum height on the route. Your job is to
write a program that finds this route.
You are given:
the number of crossings and their altitudes, and
the roads by which these crossings are connected.
Your
program must find the route that minimizes the altitude difference
between the highest and the lowest point on the route. If there are
multiple possibilities, choose the shortest one.
For example:
In
this case the shortest path from 1 to 7 would be through 2, 3 and 4,
but the altitude difference of that path is 8. So, you prefer to go
through 5, 6 and 4 for an altitude difference of 2. (Note that going
from 6 directly to 7 directly would have the same difference in
altitude, but the path would be longer!)
One
line with two integers n (1 <= n <= 100) and m (0 <= m <=
5000): the number of crossings and the number of roads. The crossings
are numbered 1..n.
n lines with one integer hi (0 <= hi <= 1 000 000 000): the altitude of the i-th crossing.
m lines with three integers aj , bj (1 <= aj , bj <= n) and cj (1 <= cj <= 1 000 000): this indicates that there is a two-way road between crossings aj and bj of length cj . You may assume that the altitude on a road between two crossings changes linearly.
You
start at crossing 1 and the contest is at crossing n. It is guaranteed
that it is possible to reach the programming contest from your home.
the minimum altitude difference, and
the length of shortest path with this altitude difference.
7 9
4
9
1
3
3
5
4
1 2 1
2 3 1
3 4 1
4 7 1
1 5 4
5 6 4
6 7 4
5 3 2
6 4 2
题意:有n个点m条边,每个点都有一个高度,问在保证高度之差最小的情况下从1点到第n点,最小高度差和最短路分别是多少?
#include <stdio.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <queue>
#include <string.h>
using namespace std;
typedef long long ll;
const int N = ;
const int INF = ;
struct Node { ///枚举高度差所需要用到的结构体
int low,high;
}node[N*N];
struct Edge{
int v,w,next;
}edge[N*N];
int head[N];
ll h[N];
int graph[N][N];
int n,m;
int cmp(Node a,Node b){
return (a.high-a.low)<(b.high-b.low);
}
bool vis[N];
int d[N];
void addEdge(int u,int v,int w,int &k){
edge[k].v = v,edge[k].w = w;
edge[k].next = head[u],head[u]=k++;
}
void spfa(int s,int low,int high){
queue<int > q;
for(int i=;i<=n;i++){
d[i] = INF;
vis[i] = false;
}
d[s] = ;
q.push(s);
while(!q.empty()){
int u = q.front();
q.pop();
vis[u] = false;
if(h[u]>high||h[u]<low) continue;
for(int k = head[u];k!=-;k=edge[k].next){
int v = edge[k].v,w = edge[k].w;
if(h[v]>high||h[v]<low) continue;
if(d[v]>d[u]+w){
d[v] = d[u]+w;
if(!vis[v]){
vis[v]=true;
q.push(v);
}
}
}
}
}
int main()
{
int tcase;
scanf("%d",&tcase);
while(tcase--){
memset(head,-,sizeof(head));
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++){
scanf("%lld",&h[i]);
}
int tot = ;
for(int i=;i<=m;i++){
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
addEdge(a,b,c,tot);
addEdge(b,a,c,tot);
}
int k = ;
for(int i=;i<=n;i++){
for(int j=i;j<=n;j++){
if(h[i]<h[j]){
node[k].low = h[i]; node[k++].high = h[j];
}
else {
node[k].low = h[j];
node[k++].high = h[i];
}
}
}
sort(node,node+k,cmp);
for(int i=;i<k;i++){
spfa(,node[i].low,node[i].high);
if(d[n]<INF){
printf("%d %d\n",node[i].high-node[i].low,d[n]);
break;
}
}
}
}
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