BZOJ2763 [JLOI2011]飞行路线(洛谷P4568)

分层图最短路

BZOJ题目传送门
洛谷题目传送门

妙蛙

把原图分成k+1层,每一层向下一层连距离为0的边,然后跑一遍最短路就好了。

代码:

#include
#include
#include
#include
#include
#define N 110005
#define F inline
using namespace std;
struct edge{ int nxt,to,d; }ed[N*20];
struct P{ int x,d; };
int n,m,k,p,s,t,ans=1e9,h[N],d[N];
priority_queue 

q; F char readc(){ static char buf[100000],*l=buf,*r=buf; if (l==r) r=(l=buf)+fread(buf,1,100000,stdin); return l==r?EOF:*l++; } F int _read(){ int x=0; char ch=readc(); while (!isdigit(ch)) ch=readc(); while (isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=readc(); return x; } #define add(x,y,z) ed[++k]=(edge){h[x],y,z},h[x]=k F bool operator < (P a,P b){ return a.d>b.d; } F int Dij(){ for (int i=1;i<=n*(p+1);i++) d[i]=1e9; for (q.push((P){s,d[s]=0});!q.empty();q.pop()){ if (q.top().d>d[q.top().x]) continue; int x=q.top().x; for (int i=h[x],v;i;i=ed[i].nxt) if (d[x]+ed[i].dfor (int i=0;i<=p;i++) ans=min(ans,d[t+n*i]); return ans; } int main(){ n=_read(),m=_read(),p=_read(),s=_read()+1,t=_read()+1; for (int i=1,x,y,z;i<=m;i++){ x=_read()+1,y=_read()+1,z=_read(),add(x,y,z),add(y,x,z); for (int j=0,u=0,v=n;j0),add(y+u,x+v,0); add(x+v,y+v,z),add(y+v,x+v,z); } } return printf("%d\n",Dij()),0; }

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