思路:
1> 求 1 到所有点的最短路径,
2> 求 n 到所有点的最长路径
3> 做差。
1> 求 1 到所有点的最短路径,
2> 求 n 到所有点的最长路径
3> 做差。
看是否可达,判断是否入过队。
#include<stdio.h> #include<queue> #include<string.h> #include<algorithm> using namespace std; #define MAXN 100004 #define INF 0x3f3f3f3f int dist1[MAXN],dist2[MAXN]; int vis1[MAXN],vis2[MAXN]; int head1[MAXN],head2[MAXN]; int top; struct Edge{ int to,next; }edge[MAXN*5]; void addedge(int u,int v){ edge[top].to=v; edge[top].next=head1[u]; head1[u]=top++; edge[top].to=u; edge[top].next=head2[v]; head2[v]=top++; } void SPFA1(int st){ memset(vis1,0,sizeof(vis1)); vis1[st]=1; queue<int>q; q.push(st); while(!q.empty()){ int u = q.front(); q.pop(); for(int i=head1[u];i!=-1;i=edge[i].next){ int v=edge[i].to; dist1[v]=min(dist1[v],dist1[u]); if(!vis1[v]){ vis1[v]=1; q.push(v); } } } } void SPFA2(int st){ memset(vis2,0,sizeof(vis2)); vis2[st]=1; queue<int>q; q.push(st); while(!q.empty()){ int u = q.front(); q.pop(); for(int i=head2[u];i!=-1;i=edge[i].next){ int v=edge[i].to; dist2[v]=max(dist2[v],dist2[u]); if(!vis2[v]){ vis2[v]=1; q.push(v); } } } } int main(){ int n,m; while(scanf("%d%d",&n,&m)!=EOF){ memset(head1,-1,sizeof(head1)); memset(head2,-1,sizeof(head2)); top=0; for(int i=1;i<=n;i++){ scanf("%d",&dist1[i]); dist2[i]=dist1[i]; } while(m--){ int a,b,c; scanf("%d%d%d",&a,&b,&c); addedge(a,b); if(c==2) addedge(b,a); } SPFA1(1); SPFA2(n); int minn=0; for(int i=1;i<=n;i++){ if(vis1[i]&&vis2[i]) minn=max(minn,dist2[i]-dist1[i]); } printf("%d\n",minn); } return 0; }