题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1598
题意：求 1-k 短路的路程，不存在输出-1
裸 K 短路，细节在于 S 变来变去.... 大体同 此题
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#include <cmath> #include <queue> #include <cstdio> #include <iomanip> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define N 1010 #define M 10010 #define ll long long using namespace std; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,m,k,S,T,tim; struct zgz { int next,to,val; }edge[M],edge2[M]; int cnt=1,cnt2=2,head[M],head2[M]; int ans[M]; void add(int from,int to,int val) { edge[cnt].to=to; edge[cnt].val=val; edge[cnt].next=head[from]; head[from]=cnt++; } void add2(int from,int to,int val) { edge2[cnt].to=to; edge2[cnt].val=val; edge2[cnt].next=head2[from]; head2[from]=cnt2++; } queue<int> Q; int d[N]; bool vis[N]; void spfa() { memset(d,0x3f,sizeof(d)); d[S]=0,vis[S]=1; Q.push(S); while(!Q.empty()) { int x=Q.front();Q.pop();vis[x]=0; for(int i=head2[x];i;i=edge2[i].next) { int to=edge2[i].to,val=edge2[i].val; if(d[to]>d[x]+val) { d[to]=d[x]+val; if(!vis[to]) { vis[to]=1; Q.push(to); } } } } } struct data { int x,g; }; bool operator < (data x,data y) { return x.g>y.g; } priority_queue<data> q; void astar() { q.push((data){S,d[S]}); while(!q.empty()) { data now=q.top();q.pop(); int x=now.x,f=now.g; f-=d[x]; if(x==T) { ans[++tim]=f; if(tim==k)return ; continue; } for(int i=head[x];i;i=edge[i].next) { int to=edge[i].to,val=edge[i].val; q.push((data){to,f+val+d[to]}); } } } int main() { n=read(),m=read(),k=read();T=1; memset(ans,-1,sizeof(ans)); for(int i=1;i<=m;i++) { int u=read(),v=read(),w=read(); add(u,v,w);add2(v,u,w); } S=1; spfa(); S=n; astar(); for(int i=1;i<=k;i++) printf("%d\n",ans[i]); } |
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