Wormholes
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 43809 | Accepted: 16085 |
Description
While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ’s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.
As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .
To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.
Input
Line 1: A single integer,
F.
F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively:
N,
M, and
W
Lines 2..
M+1 of each farm: Three space-separated numbers (
S,
E,
T) that describe, respectively: a bidirectional path between
S and
E that requires
T seconds to traverse. Two fields might be connected by more than one path.
Lines
M+2..
M+
W+1 of each farm: Three space-separated numbers (
S,
E,
T) that describe, respectively: A one way path from
S to
E that also moves the traveler back
T seconds.
Output
Lines 1..
F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).
Sample Input
2 3 3 1 1 2 2 1 3 4 2 3 1 3 1 3 3 2 1 1 2 3 2 3 4 3 1 8
Sample Output
NO YES
Hint
For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.
题意:
虫洞问题,现在有n个点,m条边,代表现在可以走的通路,比如从a到b和从b到a需要花费c时间
现在在地上出现了w个虫洞,虫洞的意义就是你从a到b话费的时间是-c(时间倒流,并且虫洞是单向的)
现在问你从某个点开始走,能回到从前
Bellman-ford
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int top;
struct node
{
int u,v,t;
}path[6010];
int n,m,w;
void add(int u,int v,int t)
{
path[top].u=u;
path[top].v=v;
path[top++].t=t;
}
int bellman_ford(int n)
{
int weight[520];
for(int i=0;i<=n;i++)
weight[i]=INF;
weight[1]=0;
int u,v,t;
for(int i=0;i<n-1;i++){
for(int j=0;j<top;j++){
u=path[j].u;
v=path[j].v;
t=path[j].t;
if(weight[u]+t<weight[v])
weight[v]=weight[u]+t;
}
}
for(int j=0;j<top;j++){
u=path[j].u;
v=path[j].v;
t=path[j].t;
if(weight[u]+t<weight[v])
return 0;
}
return 1;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&w);
top=0;
int u,v,t;
for(int i=0;i<m;i++){
scanf("%d%d%d",&u,&v,&t);
add(u,v,t);
add(v,u,t);
}
for(int i=0;i<w;i++){
scanf("%d%d%d",&u,&v,&t);
add(u,v,-t);
}
if(!bellman_ford(n))
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
SPFA
建立一个队列,初始时队列里只有起始点,再建立一个表格记录起始点到所有点的最短路径(该表格的初始值要赋为极大值,该点到他本身的路径赋为0)。然后执行松弛操作,用队列里有的点作为起始点去刷新到所有点的最短路,如果刷新成功且被刷新点不在队列中则把该点加入到队列最后。重复执行直到队列为空。
判断有无负环:
如果某个点进入队列的次数超过N次则存在负环(SPFA无法处理带负环的图)
首先建立起始点a到其余各点的
最短路径表格
首先源点a入队,当队列非空时:
1、队首元素(a)出队,对以a为起始点的所有边的终点依次进行松弛操作(此处有b,c,d三个点),此时路径表格状态为:
在松弛时三个点的最短路径估值变小了,而这些点队列中都没有出现,这些点
需要入队,此时,队列中新入队了三个结点b,c,d
队首元素b点出队,对以b为起始点的所有边的终点依次进行松弛操作(此处只有e点),此时路径表格状态为:
在最短路径表中,e的最短路径估值也变小了,e在队列中不存在,因此e也要
入队,此时队列中的元素为c,d,e
队首元素c点出队,对以c为起始点的所有边的终点依次进行松弛操作(此处有e,f两个点),此时路径表格状态为:
在最短路径表中,e,f的最短路径估值变小了,e在队列中存在,f不存在。因此
e不用入队了,f要入队,此时队列中的元素为d,e,f
队首元素d点出队,对以d为起始点的所有边的终点依次进行松弛操作(此处只有g这个点),此时路径表格状态为:
在最短路径表中,g的最短路径估值没有变小(松弛不成功),没有新结点入队,队列中元素为f,g
队首元素f点出队,对以f为起始点的所有边的终点依次进行松弛操作(此处有d,e,g三个点),此时路径表格状态为:
在最短路径表中,e,g的最短路径估值又变小,队列中无e点,e入队,队列中存在g这个点,g不用入队,此时队列中元素为g,e
队首元素g点出队,对以g为起始点的所有边的终点依次进行松弛操作(此处只有b点),此时路径表格状态为:
在最短路径表中,b的最短路径估值又变小,队列中无b点,b入队,此时队列中元素为e,b
队首元素e点出队,对以e为起始点的所有边的终点依次进行松弛操作(此处只有g这个点),此时路径表格状态为:
在最短路径表中,g的最短路径估值没变化(松弛不成功),此时队列中元素为b
队首元素b点出队,对以b为起始点的所有边的终点依次进行松弛操作(此处只有e这个点),此时路径表格状态为:
在最短路径表中,e的最短路径估值没变化(松弛不成功),此时队列为空了
最终a到g的最短路径为14
#include<queue>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 10005
int map[MAXN][MAXN];
int dis[MAXN];
int num[MAXN],vis[MAXN];
int n,m,w;
bool SPFA()
{
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
for(int i=1;i<=n;i++)
dis[i]=INF;
queue<int>q;
dis[0]=0; vis[0]=1;
q.push(0); num[0]++;
while(q.size())
{
int p=q.front();
q.pop();
vis[p]=0;
for(int i=1;i<=n;i++){
if(dis[p]+map[p][i]<dis[i]){
dis[i]=dis[p]+map[p][i];
if(vis[i]==0){
vis[i]=1;
num[i]++;
q.push(i);
if(num[i]>n)
return true;
}
}
}
}
return false;
}
int main()
{
int T;
//freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&w);
for(int i=1;i<=n;i++)
for(int j=1;j<=i;j++)
map[i][j]=map[j][i]=INF;
int a,b,c;
for(int i=1;i<=m;i++){
scanf("%d%d%d",&a,&b,&c);
if(c<map[a][b])
map[a][b]=map[b][a]=c;
}
for(int i=1;i<=w;i++){
scanf("%d%d%d",&a,&b,&c);
map[a][b]=-c;
}
for(int i=1;i<=n;i++)
map[0][i]=0;
if(SPFA())
puts("YES");
else
puts("NO");
}
return 0;
}
#include<queue>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 6000
int dis[MAXN],head[MAXN],num[MAXN],vis[MAXN];
int n,m,w;
int top;
struct node
{
int v,w,next;
}path[MAXN];
void add(int u,int v,int w)
{
path[top].v=v;
path[top].w=w;
path[top].next=head[u];
head[u]=top++;
}
bool SPFA()
{
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
for(int i=1;i<=n;i++)
dis[i]=INF;
queue<int>q;
vis[1]=1; dis[1]=0;
q.push(1); num[1]++;
while(q.size())
{
int p=q.front();
q.pop();
vis[p]=0;
for(int i=head[p];i!=-1;i=path[i].next){
int t=path[i].v;
if(dis[t]>dis[p]+path[i].w){
dis[t]=dis[p]+path[i].w;
if(vis[t]==0){
vis[t]=1;
q.push(t);
num[t]++;
if(num[t]>n)
return true;
}
}
}
}
return false;
}
int main()
{
int T;
//freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--)
{
top=0;
scanf("%d%d%d",&n,&m,&w);
memset(head,-1,sizeof(head));
int a,b,c;
for(int i=1;i<=m;i++){
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);
add(b,a,c);
}
for(int i=1;i<=w;i++){
scanf("%d%d%d",&a,&b,&c);
add(a,b,-c);
}
if(SPFA())
puts("YES");
else
puts("NO");
}
return 0;
}