POJ 3259-Wormholes (Bellman-Ford&&SPFA) （模板题）

Wormholes

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 39754 Accepted: 14602

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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.

先上Bellman-Ford算法。

借鉴自：http://blog.csdn.net/niushuai666/article/details/6791765

算法非常简单。

第一，初始化所有点。每一个点保存一个值，表示从原点到达这个点的距离，将原点的值设为0，其它的点的值设为无穷大（表示不可达）。
第二，进行循环，循环下标为从1到n－1（n等于图中点的个数）。在循环内部，遍历所有的边，进行松弛计算。
第三，遍历途中所有的边（edge（u，v）），判断是否存在这样情况：
d（v） > d (u) + w(u,v)
若存在则返回false，表示途中存在从源点可达的权为负的回路。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdlib>#include <string>#include <vector>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>using namespace std;#define INF 0x3f3f3f3const int N=100005;const int mod=1e9+7;int f,n,m,w;int d[505];struct Edge{    int u,v,cost;}edge[N];bool bellman_ford(){    for (int i=1; i<n; i++) {        for (int j=1; j<=2*m+w; j++) {            if (d[edge[j].v]>d[edge[j].u]+edge[j].cost) {                d[edge[j].v]=d[edge[j].u]+edge[j].cost;            }        }    }    bool flag=true;    for (int j=1; j<=2*m+w; j++)        if (d[edge[j].v]>d[edge[j].u]+edge[j].cost) {            flag=false;            break;        }    return flag;}int main(){    int s,e,t;    cin>>f;    while (f--) {        cin>>n>>m>>w;        memset(d, INF, sizeof(d));        for (int i=1; i<=2*m; i+=2) {            scanf("%d %d %d",&edge[i].u,&edge[i].v,&edge[i].cost);            edge[i+1].u=edge[i].v;            edge[i+1].v=edge[i].u;            edge[i+1].cost=edge[i].cost;        }        for (int i=1; i<=w; i++) {            scanf("%d %d %d",&edge[i+2*m].u,&edge[i+2*m].v,&edge[i+2*m].cost);            edge[i+2*m].cost*=-1;        }        if (!bellman_ford()) {            cout<<"YES\n";        }        else {            cout<<"NO\n";        }    }    return 0;}

下面是SPFA算法：http://baike.baidu.com/link?url=FNAoppxXYYnhHZI1MGJnZTs9-2MBZPG08RkqHEFWYHDUfrDW2zYd0AjfxthdCcUqM6FrvWEcr90mKTvJ1Gk_jqu9O3aRS5dm85d3-_cI_OzVocyLMGdY7p78E5IXiiheVsMHzu6CFD_dgOJP2xpgMa

找的百度百科资料，照本宣科的用，感觉和bellman-ford算法大意相同。

我们用数组d记录每个结点的最短路径估计值，而且用邻接表来存储图G。我们采取的方法是动态逼近法：设立一个先进先出的队列用来保存待优化的结点，优化时每次取出队首结点u，并且用u点当前的最短路径估计值对离开u点所指向的结点v进行松弛操作，如果v点的最短路径估计值有所调整，且v点不在当前的队列中，就将v点放入队尾。这样不断从队列中取出结点来进行松弛操作，直至队列空为止。若有节点放入队列的次数大于n（结点个数），那么就存在负环。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdlib>#include <string>#include <vector>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <map>#include <set>using namespace std;#define INF 0x3f3f3f3const int N=100005;const int mod=1e9+7;int f,n,m,w;int d[505],cnt[505];vector<pair<int, int> > a[505];bool in_queue[505];queue<int> q;bool SPFA(int st){    memset(d, 0x3f, sizeof(d));    memset(in_queue, 0, sizeof(in_queue));    memset(cnt, 0, sizeof(cnt));    d[st]=0;    q.push(st);    in_queue[st]=1;    cnt[st]=1;    while (!q.empty()) {        int u=q.front();        if (cnt[u]>n) {            return false;        }        for (int i=0; i<a[u].size(); i++) {            pair<int, int> v=a[u][i];            if (d[u]+v.second<d[v.first]) {                d[v.first]=d[u]+v.second;                if (!in_queue[v.first]) {                    q.push(v.first);                    in_queue[v.first]=1;                    cnt[v.first]++;                }            }        }        q.pop();        in_queue[u]=0;    }    return true;}int main(){    int s,e,t;    cin>>f;    while (f--) {        cin>>n>>m>>w;        while (!q.empty()) {            q.pop();        }        for (int i=0; i<=n; i++) {            a[i].clear();        }        for (int i=1; i<=m; i++) {            scanf("%d %d %d",&s,&e,&t);            a[s].push_back(make_pair(e, t));            a[e].push_back(make_pair(s, t));        }        for (int i=1; i<=w; i++) {            scanf("%d %d %d",&s,&e,&t);            a[s].push_back(make_pair(e, -t));        }        int flag=1;        for (int i=1; i<=n; i++) {            if (!SPFA(i)) {                flag=0;                break;            }        }        if (flag) {            cout<<"NO\n";        } else {            cout<<"YES\n";        }    }    return 0;}