URAL 1072 Routing(最短路)
Routing
Memory limit: 64 MB
- Each computer has one or more net interfaces.
- Each interface is identified by its IP-address and a subnet mask —
these are two four-byte numbers with a point after each byte. A subnet
mask has a binary representation as follows: there are k 1-bits, then — m 0-bits, k+m=8*4=32 (e.g., 212.220.35.77 — is an IP-address and 255.255.255.128 — is a subnet mask). - Two computers belong to the same subnet, if and only if (IP1 AND NetMask1) = (IP2 AND NetMask2), where IPi and NetMaski — are an IP-address and subnet mask of i-th computer, AND — is bitwise.
- A packet is transmitted between two computers of one subnet directly.
- If two computers belong to different subnets, a packet is to be
transmitted via some other computers. The packet can pass from one
subnet to another only on computer that has both subnets interfaces.
Input
lines — descriptions of the interfaces, i.e. its IP-address and a
subnet mask. The last line of an input contains two integers — the
numbers of the computers that you are to find a way between them.
Output
word “Yes” if the route exists, then in the next line the computer
numbers passed by the packet, separated with a space. The word “No”
otherwise.
Sample
input | output |
---|---|
6 |
Yes |
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
typedef long long ll;
using namespace std;
const int N = ;
const int M = ;
long long yu[N][];
int map[N][N];
int len[N];
int a[N],n;
bool check(int x,int y) {
for(int i=; i<a[x]; i++)
for(int k=; k<a[y]; k++)
if( yu[x][i] == yu[y][k] )
return true;
return false;
}
int pre[N];
void output(int t) {
if( pre[t] == - ) return;
output(pre[t]);
printf(" %d",t);
}
void Dijkstra(int s,int t) {
bool used[N];
int dis[N];
memset(used,false,sizeof(used));
fill(dis,dis+N,INT_MAX);
memset(pre,-,sizeof(pre));
dis[s] = ;
used[s] = true;
int now = s;
for(int i=; i<n; i++) {
for(int k=; k<=n; k++)
if( map[now][k] && dis[k] > dis[now] + ) {
dis[k] = dis[now] + ;
pre[k] = now;
}
int mmin = INT_MAX;
for(int k=; k<=n; k++)
if( !used[k] && dis[k] < mmin )
mmin = dis[now = k];
used[now] = ;
}
if( dis[t] == INT_MAX ) {
printf("No\n");
return ;
}
printf("Yes\n");
printf("%d",s);
output(t);
printf("\n");
}
int main() {
int t1[],t2[],s,t;
scanf("%d",&n);
memset(len,,sizeof(len));
memset(yu,,sizeof(yu));
memset(map,,sizeof(map));
for(int i=; i<=n; i++) {
scanf("%d",&a[i]);
for(int k=; k<a[i]; k++) {
scanf("%d.%d.%d.%d",&t1[],&t1[],&t1[],&t1[]);
scanf("%d.%d.%d.%d",&t2[],&t2[],&t2[],&t2[]);
for(int p=; p<; p++) {
yu[i][k] *= ;
yu[i][k] += ( t1[p] & t2[p] );
}
}
for(int k=; k<i; k++)
if( check(k,i) )
map[i][k] = map[k][i] = ;
}
scanf("%d %d",&s,&t);
Dijkstra(s,t); return ;
}
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