danihao123's Blog

[BZOJ 2763]飞行路线

分层图最短路处女作TAT

很明显要求你求一个分层图上的最短路。不过没必要把分层图构出来，手写转移即可。还有卡SPFA是怎么回事？出题人你粗来我保证不打死你……

然而沉迷pb_ds，不能自拔。

代码：

/**************************************************************
    Problem: 2763
    User: danihao123
    Language: C++
    Result: Accepted
    Time:268 ms
    Memory:3136 kb
****************************************************************/
 
#include <cstdio>
#include <queue>
#include <algorithm>
#include <utility>
#include <cstring>
#include <cctype>
#include <ext/pb_ds/priority_queue.hpp>
using namespace std;
const int maxn=10001,maxm=100001,maxk=11;
typedef pair<int,int> pii;
int first[maxn];
int next[maxm],to[maxm],dist[maxm];
int graph_cnt=0;
inline void Add_Edge(int u,int v,int d){
    graph_cnt++;
    next[graph_cnt]=first[u];
    first[u]=graph_cnt;
    to[graph_cnt]=v;
    dist[graph_cnt]=d;
}
 
struct Node{
    pii pnt;
    int d;
    bool operator <(const Node& x) const{
        return d<x.d;
    }
    bool operator >(const Node& x) const{
        return d>x.d;
    }
};
 
typedef __gnu_pbds::priority_queue<Node,greater<Node> > Heap;
Heap::point_iterator ite[maxn][maxk];
Heap Q;
int n,k;
bool vis[maxn][maxk];
int d[maxn][maxk];
inline void relax(int u,int v,int di){
    d[u][v]=di;
    if(ite[u][v]!=0)
        Q.modify(ite[u][v],(Node){make_pair(u,v),di});
    else
        ite[u][v]=Q.push((Node){make_pair(u,v),di});
}
int Dijkstra(int s,int t){
    register int i,u,v;
    pii temp;
    memset(d,0x7f,sizeof(d));
    d[s][0]=0;
    ite[s][0]=Q.push((Node){make_pair(s,0),0});
    while(!Q.empty()){
        temp=Q.top().pnt;
        Q.pop();
        u=temp.first;
        v=temp.second;
        if(vis[u][v])
            continue;
        vis[u][v]=true;
        for(i=first[u];i;i=next[i]){
            if(v<k){
                if(d[u][v]<d[to[i]][v+1]){
                    relax(to[i],v+1,d[u][v]);
                }
            }
            if(d[u][v]+dist[i]<d[to[i]][v]){
                relax(to[i],v,d[u][v]+dist[i]);
            }
        }
    }
    register int ans=0x7f7f7f7f;
    for(i=0;i<=k;i++)
        ans=min(ans,d[t][i]);
    return ans;
}
int main(){
    int m,s,t,u,v,d;
    register int i;
    scanf("%d%d%d%d%d",&n,&m,&k,&s,&t);
    for(i=1;i<=m;i++){
        scanf("%d%d%d",&u,&v,&d);
        Add_Edge(u,v,d);
        Add_Edge(v,u,d);
    }
    printf("%d\n",Dijkstra(s,t));
    return 0;
}

[BZOJ 4152]The Captain

这题初看有些棘手。但是毕竟是最短路，[tex]min(abs(x_{1}-x_{2}),abs(y_{1}-y_{2}))[/tex]这样的边，大可以分成两条边来建，于是乎min就形同虚设了。并且我们可以看到这样建图有一个好处：跨越若干点的方案可以一定可以分成图上的若干边。问题大为简化，迎刃而解。

但这提有个丧病的地方：卡SPFA。加了一些优化照样挂。所以我一气之下写了可并堆优化的Dijkstra :)当然可并堆用的是pb_ds辣。

代码：

/**************************************************************
    Problem: 4152
    User: danihao123
    Language: C++
    Result: Accepted
    Time:4888 ms
    Memory:17412 kb
****************************************************************/
 
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <utility>
#include <ext/pb_ds/priority_queue.hpp>
#include <cctype>
#include <bitset>
#ifdef DEBUG
#include <cassert>
#endif
using namespace std;
const int maxn=200001;
int n;
inline int abs(int x){
    return x<0?-x:x;
}
  
int first[maxn];
int next[maxn*4],to[maxn*4],dist[maxn*4];
int graph_cnt=0;
inline void Add_Edge(int x,int y,int d){
    graph_cnt++;
    next[graph_cnt]=first[x];
    first[x]=graph_cnt;
    to[graph_cnt]=y;
    dist[graph_cnt]=d;
}
  
int d[maxn];
bitset<maxn> vis;
typedef pair<int,int> my_pair;
typedef __gnu_pbds::priority_queue<my_pair,greater<my_pair> > Heap;
Heap::point_iterator ite[maxn];
Heap Q;
int dij(){
    register int i,u;
    memset(d,0x7f,sizeof(d));
    d[1]=0;
    ite[1]=Q.push(make_pair(0,1));
    while(!Q.empty()){
        u=Q.top().second;
        Q.pop();
        if(vis[u])
            continue;
        vis[u]=true;
        for(i=first[u];i;i=next[i]){
            if(d[to[i]]>(dist[i]+d[u])){
                d[to[i]]=dist[i]+d[u];
                if(ite[to[i]]!=0)
                    Q.modify(ite[to[i]],make_pair(d[to[i]],to[i]));
                else
                    ite[to[i]]=Q.push(make_pair(d[to[i]],to[i]));
            }
        }
    }
    return d[n];
}
  
int pr[maxn][2];
int order1[maxn],order2[maxn];
int cmp1(const int i,const int j){
    return pr[i][0]<pr[j][0];
}
int cmp2(const int i,const int j){
    return pr[i][1]<pr[j][1];
}
// I/O优化
inline int readint(){
    char c=getchar();
    register int x=0;
    while(!isdigit(c))
        c=getchar();
    while(isdigit(c)){
        x=x*10+c-'0';
        c=getchar();
    }
    return x;
}
int main(){
    register int i;
    n=readint();
    for(i=1;i<=n;i++){
        pr[i][0]=readint();
        pr[i][1]=readint();
        order1[i]=i;
        order2[i]=i;
    }
    sort(order1+1,order1+1+n,cmp1);
    sort(order2+1,order2+1+n,cmp2);
    for(i=1;i<=n;i++){
        if(i!=1){
            Add_Edge(order1[i],order1[i-1],pr[order1[i]][0]-pr[order1[i-1]][0]);
            Add_Edge(order2[i],order2[i-1],pr[order2[i]][1]-pr[order2[i-1]][1]);
        }
        if(i!=n){
            Add_Edge(order1[i],order1[i+1],pr[order1[i+1]][0]-pr[order1[i]][0]);
            Add_Edge(order2[i],order2[i+1],pr[order2[i+1]][1]-pr[order2[i]][1]);
        }
    }
    printf("%d\n",dij());
    return 0;
}