本文主要是介绍路障【SPFA】,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
题目大意:
求一个无向图的次短路。
Input I n p u t
4 4
1 2 100
2 4 200
2 3 250
3 4 100
Output O u t p u t
450
思路:
这道题,正解是跑两遍 SPFA S P F A ,一遍是从点 1 1 ,求出到达其他点的最短路径,记作;再从点 n n ,也求出到达其他点的最短路径,记作;最后枚举每条边,若两个端点分别是 u[i] u [ i ] 和 v[i] v [ i ] ,长度为 dis[i] d i s [ i ] ,那么就是求次小的
dis[i]+dis1[u[i]]+dis2[v[i]] d i s [ i ] + d i s 1 [ u [ i ] ] + d i s 2 [ v [ i ] ]
或
dis[i]+dis2[u[i]]+dis1[v[i]] d i s [ i ] + d i s 2 [ u [ i ] ] + d i s 1 [ v [ i ] ]
代码:
#include <cstdio>
#include <iostream>
#include <queue>
#define fre(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
using namespace std;const int Inf=99999999;
int n,m,x,y,z,t,dis1[30001],dis2[30001],vis[30001],head[30001],dis[300001],X[300001],Y[300001],ans,minn,sum;struct edge //邻接表
{int next,to,dis;
}e[300001];void add(int from,int to,int d) //连边
{t++;e[t].to=to;e[t].dis=d;e[t].next=head[from];head[from]=t;
}void spfa1() //从点1开始跑SPFA
{queue<int> q;for (int i=1;i<=n;i++){vis[i]=0;dis1[i]=Inf;}vis[1]=1;dis1[1]=0;q.push(1);while (q.size()) //队列不为空{int u=q.front();vis[u]=0;q.pop();for (int i=head[u];i;i=e[i].next){int v=e[i].to;if (dis1[v]>dis1[u]+e[i].dis) //更新最短路{dis1[v]=dis1[u]+e[i].dis;if (!vis[v]){q.push(v);vis[v]=1;}}}}
}void spfa2() //从点n开始跑SPFA
{queue<int> q;for (int i=1;i<=n;i++){vis[i]=0;dis2[i]=Inf;}vis[n]=1;dis2[n]=0;q.push(n);while (q.size()){int u=q.front();vis[u]=0;q.pop();for (int i=head[u];i;i=e[i].next){int v=e[i].to;if (dis2[v]>dis2[u]+e[i].dis){dis2[v]=dis2[u]+e[i].dis;if (!vis[v]){q.push(v);vis[v]=1;}}}}
}int main()
{fre(block);scanf("%d%d",&n,&m);for (int i=1;i<=m;i++){scanf("%d%d%d",&x,&y,&z);dis[i]=z;X[i]=x;Y[i]=y; //记录add(x,y,z);add(y,x,z); //连边}spfa1();spfa2();minn=ans=Inf;for (int i=1;i<=m;i++) //求次短路{sum=dis[i]+dis1[X[i]]+dis2[Y[i]];if (sum<minn){ans=minn;minn=sum;}else if (sum<ans&&sum!=minn) ans=sum;sum=dis[i]+dis2[X[i]]+dis1[Y[i]];if (sum<minn){ans=minn;minn=sum;}else if (sum<ans&&sum!=minn) ans=sum;}printf("%d\n",ans);return 0;
}
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