最短路径 专题总结
Posted
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了最短路径 专题总结相关的知识,希望对你有一定的参考价值。
一.模板:
1.dijsktra算法(矩阵):
1 int n,cost[1005][1005],dis[1005],vis[1005],dp[1005]; 2 void dijkstra(int st) 3 { 4 memset(vis,0,sizeof(vis)); 5 for(int i = 1; i<=n; i++) 6 dis[i] = (i==st?0:INF); 7 8 for(int i = 1; i<=n; i++) 9 { 10 int k, minn = INF; 11 for(int j = 1; j<=n; j++) 12 if(!vis[j]&&dis[j]<minn) 13 minn = dis[k=j]; 14 15 vis[k] = 1; 16 for(int j = 1; j<=n; j++) 17 if(!vis[j] && cost[k][j]!=INF) 18 dis[j] = min(dis[j], dis[k] + cost[k][j]); 19 } 20 }
2.dijsktra算法(前向星):
1 struct edge 2 { 3 int to, w, next; 4 }edge[maxn*maxn]; 5 int cnt, head[maxn]; 6 7 void addedge(int u, int v, int w) 8 { 9 edge[cnt].to = v; 10 edge[cnt].w = w; 11 edge[cnt].next = head[u]; 12 head[u] = cnt++; 13 } 14 15 void init() 16 { 17 cnt = 0; 18 memset(head, -1, sizeof(head)); 19 } 20 21 int dis[maxn], vis[maxn]; 22 void dijkstra(int st) 23 { 24 memset(vis, 0, sizeof(vis)); 25 for(int i = 1; i<=n; i++) 26 dis[i] = (i==st?0:INF); 27 28 for(int i = 1; i<=n; i++) 29 { 30 int k, minn = INF; 31 for(int j = 1; j<=n; j++) 32 if(!vis[j] && dis[j]<minn) 33 minn = dis[k=j]; 34 35 vis[k] = 1; 36 for(int j = head[k]; j!=-1; j = edge[j].next) 37 if(!vis[edge[j].to]) 38 dis[edge[j].to] = min(dis[edge[j].to], dis[k] + edge[j].w); 39 } 40 }
3.bellman-ford算法(前向星):
1 struct edge 2 { 3 int u, v, w; 4 }edge[maxn*maxn]; 5 int cnt; 6 7 void addedge(int u, int v, int w) 8 { 9 ++cnt; 10 edge[cnt].u = u; 11 edge[cnt].v = v; 12 edge[cnt].w = w; 13 } 14 15 int dis[maxn]; 16 bool bellman(int st) 17 { 18 for(int i = 1; i<=n; i++) 19 dis[i] = (i==st?0:INF); 20 21 for(int i = 1; i<=n-1; i++) 22 { 23 for(int j = 1; j<=cnt; j++) 24 { 25 int u = edge[j].u; 26 int v = edge[j].v; 27 if(dis[u]!=INF) 28 dis[v] = min(dis[v], dis[u]+edge[j].w); 29 } 30 } 31 32 for(int i = 1; i<=cnt; i++) //判断是否存在负环 33 { 34 int u = edge[i].u; 35 int v = edge[i].v; 36 if(dis[u]!=INF && dis[u]+edge[i].w<dis[v]) 37 return false; 38 } 39 return true; 40 }
4.spfa算法(前向星):
1 struct edge 2 { 3 int to, w, next; 4 }edge[maxn*maxn]; 5 int cnt, head[maxn]; 6 7 void addedge(int u, int v, int w) 8 { 9 edge[cnt].to = v; 10 edge[cnt].w = w; 11 edge[cnt].next = head[u]; 12 head[u] = cnt++; 13 } 14 15 void init() 16 { 17 cnt = 0; 18 memset(head, -1, sizeof(head)); 19 } 20 21 int dis[maxn], times[maxn], inq[maxn]; 22 bool spfa(int st) 23 { 24 memset(inq, 0, sizeof(inq)); 25 memset(times, 0, sizeof(times)); 26 for(int i = 1; i<=n; i++) 27 dis[i] = (i==st?0:INF); 28 29 queue<int>Q; 30 Q.push(st); 31 inq[st] = 1; 32 while(!Q.empty()) 33 { 34 int u = Q.front(); 35 Q.pop(); inq[u] = 0; 36 for(int i = head[u]; i!=-1; i = edge[i].next) 37 { 38 int v = edge[i].to; 39 if(dis[v]>dis[u]+edge[i].w) 40 { 41 dis[v] = dis[u]+edge[i].w; 42 if(!inq[v]) 43 { 44 Q.push(v); 45 inq[v] = 1; 46 if(++times[v]>n) return false; 47 } 48 } 49 } 50 } 51 return true; 52 }
5.floyd算法:
1 for(int i = 1; i<=n; i++) 2 for(int j = 1; j<=n; j++) 3 scanf("%d",&dis[i][j]); 4 5 for(int k = 1; k<=n; k++) 6 for(int i = 1; i<=n; i++) 7 for(int j = 1; j<=n; j++) 8 dis[i][j] = min(dis[i][j], dis[i][k]+dis[k][j]);
二.题目类型:
1.普通最短路:
2.最短路变形:
3.求负环或正环:
4.传递闭包:
5.差分约束:
以上是关于最短路径 专题总结的主要内容,如果未能解决你的问题,请参考以下文章