POJ 3259-Wormholes (Bellman-Ford&&SPFA) （模板题）
Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total 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.
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.
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。
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 0x3f3f3f3
const 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 0x3f3f3f3
const 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;
}