bzoj2763 [JLOI2011]飞行路线(分层图最短路)

d[i][j]表示到i点，用了j次免费的最短路,每个j对应了一层。因此唤做分层图。更新时，本层自然要做，还可以跨维度的把下一层也影响了。不过我跑得好慢。。。看各位神犇都是1S以内的。。我知道了，是spfa慢。。Dijkstra才240ms。

spfa版 7380ms

#include <bits/stdc++.h>using namespace std;#define pa pair<int,int>#define ll long long#define inf 0x3f3f3f3f#define N 10010#define M 50010inline 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,h[N],num=0,d[N][12],ans=inf;//d[i][j]表示到i点，用了j次免费的最短路 bool inq[N][12];struct edge{    int to,next,val;}data[M<<1];void spfa(){    queue<pa>q;memset(d,0x3f,sizeof(d));    q.push(make_pair(s,0));inq[s][0]=1;d[s][0]=0;    while(!q.empty()){        int x=q.front().first,kx=q.front().second;q.pop();inq[x][kx]=0;        for(int i=h[x];i;i=data[i].next){            int y=data[i].to;            if(d[x][kx]+data[i].val<d[y][kx]){//转移本层                 d[y][kx]=d[x][kx]+data[i].val;                if(!inq[y][kx]) q.push(make_pair(y,kx)),inq[y][kx]=1;            }            if(kx+1<=k&&d[x][kx]<d[y][kx+1]){//转移下层                 d[y][kx+1]=d[x][kx];                if(!inq[y][kx+1]) q.push(make_pair(y,kx+1)),inq[y][kx+1]=1;            }        }    }}int main(){//  freopen("a.in","r",stdin);    n=read();m=read();k=read();s=read();t=read();    while(m--){        int x=read(),y=read(),val=read();        data[++num].to=y;data[num].next=h[x];h[x]=num;data[num].val=val;        data[++num].to=x;data[num].next=h[y];h[y]=num;data[num].val=val;    }    spfa();    for(int i=0;i<=k;++i) ans=min(ans,d[t][i]);    printf("%d\n",ans);    return 0;}

Dijkstra版 240ms

#include <bits/stdc++.h>using namespace std;#define pa pair<int,pair<int,int> >#define ll long long#define inf 0x3f3f3f3f#define N 10010#define M 50010inline 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,h[N],num=0,d[N][12],ans=inf;//d[i][j]表示到i点，用了j次免费的最短路 bool f[N][12];struct edge{    int to,next,val;}data[M<<1];void Dijkstra(){    priority_queue<pa,vector<pa>,greater<pa> >q;    memset(d,0x3f,sizeof(d));    q.push(make_pair(0,make_pair(s,0)));d[s][0]=0;    while(!q.empty()){        int x=q.top().second.first,kx=q.top().second.second;        q.pop();if(f[x][kx]) continue;f[x][kx]=1;        for(int i=h[x];i;i=data[i].next){            int y=data[i].to;            if(d[x][kx]+data[i].val<d[y][kx]){                d[y][kx]=d[x][kx]+data[i].val;                q.push(make_pair(d[y][kx],make_pair(y,kx)));            }            if(kx+1<=k&&d[x][kx]<d[y][kx+1]){                d[y][kx+1]=d[x][kx];                q.push(make_pair(d[y][kx+1],make_pair(y,kx+1)));            }        }    }}int main(){//  freopen("a.in","r",stdin);    n=read();m=read();k=read();s=read();t=read();    while(m--){        int x=read(),y=read(),val=read();        data[++num].to=y;data[num].next=h[x];h[x]=num;data[num].val=val;        data[++num].to=x;data[num].next=h[y];h[y]=num;data[num].val=val;    }    Dijkstra();    for(int i=0;i<=k;++i) ans=min(ans,d[t][i]);    printf("%d\n",ans);    return 0;}