POJ 3259 Wormholes （判负环）

Wormholes

题目链接：

http://acm.hust.edu.cn/vjudge/contest/122685#problem/F

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： 平面上的N个点，有M条正常的道路，有W条长度为负的道路. 求是否存在负环.
##题解： 以下分别用bellman-ford和spfa实现.
##代码： bellman-ford： ``` cpp #include #include #include #include #include #define mid(a,b) ((a+b)>>1) #define LL long long #define maxn 6000 #define inf 0x3f3f3f3f #define IN freopen("in.txt","r",stdin); using namespace std;

int m,n,k;
int edges, u[maxn], v[maxn], w[maxn];
int first[maxn], next[maxn];
int dis[maxn];

void add_edge(int s, int t, int val) {
u[edges] = s; v[edges] = t; w[edges] = val;
next[edges] = first[s];
first[s] = edges++;
}

bool bellman(int s) {
for(int i=1; i<=n; i++) dis[i]=inf; dis[s] = 0;

//for(int i=1; i<n; i++) {
for(int i=1; i<=n; i++) {
    for(int e=0; e<edges; e++) if(dis[v[e]] > dis[u[e]]+w[e]){
        dis[v[e]] = dis[u[e]] + w[e];
        if(i == n) return 0;
    }
}

// for(int e=0; e<edges; e++)
// if(dis[v[e]] > dis[u[e]]+w[e])
// return 0;

return 1;

}

int main(int argc, char const *argv[])
{
//IN;

int t; cin >> t;
while(t--)
{
    cin >> n >> m >> k;
    edges = 0;
    for(int i=1; i<=m; i++) {
        int u,v,w; scanf("%d %d %d",&u,&v,&w);
        add_edge(u,v,w);
        add_edge(v,u,w);
    }
    for(int i=1; i<=k; i++) {
        int u,v,w; scanf("%d %d %d",&u,&v,&w);
        add_edge(u,v,-w);
    }

    bool flag = bellman(1);

    if(!flag) puts("YES");
    else puts("NO");
}

return 0;

}

spfa：

include

include

include

include

include

include

define mid(a,b) ((a+b)>>1)

define LL long long

define maxn 6000

define inf 0x3f3f3f3f

define IN freopen("in.txt","r",stdin);

using namespace std;

int m,n,k;
int edges, u[maxn], v[maxn], w[maxn];
int first[maxn], next[maxn];
int dis[maxn];

void add_edge(int s, int t, int val) {
u[edges] = s; v[edges] = t; w[edges] = val;
next[edges] = first[s];
first[s] = edges++;
}

queue q;
bool inq[maxn];
int inq_cnt[maxn];
bool spfa(int s) {
memset(inq, 0, sizeof(inq));
memset(inq_cnt, 0, sizeof(inq_cnt));
for(int i=1; i<=n; i++) dis[i] = inf; dis[s] = 0;
while(!q.empty()) q.pop();
q.push(s); inq_cnt[s]++;

while(!q.empty()) {
    int p = q.front(); q.pop();
    inq[p] = 0;
    for(int e=first[p]; e!=-1; e=next[e]) if(dis[v[e]] > dis[p]+w[e]){
        dis[v[e]] = dis[p] + w[e];
        if(!inq[v[e]]) {
            q.push(v[e]);
            inq[v[e]] = 1;
            inq_cnt[v[e]]++;
            if(inq_cnt[v[e]] >= n) return 0;
        }
    }
}

return 1;

}

int main(int argc, char const *argv[])
{
//IN;

int t; cin >> t;
while(t--)
{
    cin >> n >> m >> k;
    edges = 0;
    memset(first, -1, sizeof(first));
    for(int i=1; i<=m; i++) {
        int u,v,w; scanf("%d %d %d",&u,&v,&w);
        add_edge(u,v,w);
        add_edge(v,u,w);
    }
    for(int i=1; i<=k; i++) {
        int u,v,w; scanf("%d %d %d",&u,&v,&w);
        add_edge(u,v,-w);
    }

    bool flag = spfa(1);

    if(!flag) puts("YES");
    else puts("NO");
}

return 0;

}