poj 3259 Wormholes(判斷負環)
Time Limit: 2000MS | Memory Limit: 65536K |
Total Submissions: 44782 | Accepted: 16496 |
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
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
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.
Source
題意:
有n個點,m條普通的雙向的路,w條單向的蟲洞,問是否能從某個點出發走一些路再回來時回到過去。
即判斷是否有負環。
這題的資料很弱,一開始用spfa去判斷與點1連通的路中是否有負環,按理應該是wa....結果過了....
題中並沒說是連通圖,可能與1不連通的一些點也能產生負環。
如果用spfa去判斷非連通圖中是否有負環應該遍歷每個頂點去判斷是否有負環, 這樣複雜度比較高。
但是Bellman-Ford可以避開這問題,因為Bellman-Ford是去拿每一條邊鬆弛,而spfa只是用與原點連通的邊去鬆弛。
spfa程式碼:
//C++ TLE, G++ 1000+ms
#include<queue>
#include<cstring>
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn = 50005;
const int INF = 0x3f3f3f3f;
int n, m, w, k, head[maxn], dis[maxn], cnt[maxn];
bool book[maxn];
struct node
{
int v, w, next;
}edge[maxn];
void addEdge(int u, int v, int w)
{
edge[k].v = v;
edge[k].w = w;
edge[k].next = head[u];
head[u] = k;
k++;
}
bool spfa(int u)
{
memset(book, 0, sizeof(book));
memset(cnt, 0, sizeof(cnt));
for(int i = 1; i <= n; i++) dis[i] = INF;
dis[u] = 0;
cnt[u]++;
queue<int> q;
q.push(u);
while(!q.empty())
{
u = q.front(); q.pop();
book[u] = 0;
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].v, w = edge[i].w;
if(dis[u]+w < dis[v])
{
dis[v] = dis[u]+w;
if(!book[v])
{
book[v] = 1;
q.push(v);
cnt[v]++;
if(cnt[v] >= n) return 1;
}
}
}
}
return 0;
}
int main(void)
{
int t;
cin >> t;
while(t--)
{
memset(head, -1, sizeof(head));
k = 0;
scanf("%d%d%d", &n, &m, &w);
for(int i = 1; i <= m+w; i++)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
if(i > m) w = -w;
addEdge(u, v, w);
if(i <= m) addEdge(v, u, w);
}
bool flag = 0;
for(int i = 1; i <= n; i++)
{
if(spfa(i))
{
flag = 1;
break;
}
}
puts(flag ? "YES" : "NO");
}
return 0;
}
Bellman-Ford程式碼:
//47ms
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 50005;
int n, w, m, k, dis[maxn];
struct node
{
int u, v, w;
node() {}
node(int uu, int vv, int ww): u(uu), v(vv), w(ww) {}
}edge[maxn];
bool Bellman_Ford(int s)
{
for(int i = 1; i <= n; i++) dis[i] = INF;
dis[s] = 0;
for(int i = 0; i < n; i++)
{
int flag = 1;
for(int j = 0; j < k; j++)
{
int u = edge[j].u;
int v = edge[j].v;
int w = edge[j].w;
if(dis[u]+w < dis[v])
{
flag = 0;
dis[v] = dis[u]+w;
if(i == n-1) return true;
}
}
if(flag) return false;
}
return false;
}
int main(void)
{
int t;
cin >> t;
while(t--)
{
k = 0;
scanf("%d%d%d", &n, &m, &w);
for(int i = 1; i <= m+w; i++)
{
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
if(i <= m)
{
edge[k++] = node(u, v, w);
edge[k++] = node(v, u, w);
}
else edge[k++] = node(u, v, -w);
}
puts(Bellman_Ford(1) ? "YES" : "NO");
}
return 0;
}