BZOJ 4289(PA2012 Tax-最短路)

给出一个N个点M条边的无向图，经过一个点的代价是进入和离开这个点的两条边的边权的较大值，求从起点1到点N的最小代价。起点的代价是离开起点的边的边权，终点的代价是进入终点的边的边权
N<=100000
M<=200000

经典做法：把无向边拆成2条有向边(u,v,w)，边变点建图，(u,v,w1)向(v,l,w2)连max(w1,w2)的边，但是这样边数太多。

我们对于每个点把它的出边按权值排序。对于相邻2条边，小的向大的边连边，边权为边权的差值，大的向小的连边，边权为0.
然后求1点的出边到n点的出边的最短路。
注意处理一下起点。

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>#include<iomanip> #include<vector>#include<string>#include<queue>#include<stack>#include<map>#include<sstream>#include<complex>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (1000000000000000LL)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define vpi vector<pi >#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (410000+10)struct Edge{    ll from,to,dist;};struct HeapNode {    ll d,u;    bool operator< (const HeapNode& rhs) const {        return d > rhs.d;    }}; struct Dijkstra {    int n,m;    vector<Edge> edges;    vector<int> G[MAXN];    bool done[MAXN];    ll d[MAXN];    int p[MAXN]; //×î¶Ì·ÖÐÉÏÒ»Ìõ±ß    int pnode[MAXN];    void addedge(int u,int v,ll w){        edges.pb((Edge){u,v,w});        G[u].pb(m++);//      cout<<u<<' '<<v<<' '<<w<<endl;    }    void addedge2(int u,int v,ll w) {        addedge(u,v,w);addedge(v,u,w);    }    void init(int _n){        n=_n; m = 0;        Rep(i,n) G[i].clear();        edges.clear();     }     void dijkstra(int s) {        priority_queue<HeapNode> Q;        Rep(i,n) d[i]=1e15,pnode[i]=-1;        d[s]=0;        MEM(done)        Q.push((HeapNode){0,s});        while(!Q.empty()) {            HeapNode x=Q.top(); Q.pop();            int u=x.u;            if (done[u]) continue;            done[u]=1;            int mm=G[u].size();            Rep(i,mm) {                Edge e = edges[G[u][i]];                if (d[e.to]>d[u]+e.dist) {                    d[e.to]=d[u]+e.dist;                    p[e.to]=G[u][i];                    pnode[e.to]=u;                    Q.push((HeapNode){d[e.to],e.to});                }            }        }    } }S1;vector< pi  > e[MAXN];int main(){//  freopen("bzoj4289.in","r",stdin);    int n,m,cnt=0;    cin>>n>>m;    S1.init(1+(m<<1));    For(i,m) {        int u=read(),v=read(),w=read();        e[u].pb(mp(w,++cnt));        e[v].pb(mp(w,++cnt));        S1.addedge2(cnt,cnt-1,w);    }       For(i,n) sort(ALL(e[i]));    For(i,n) {        for(int j=0;j+1<e[i].size();j++) {            S1.addedge(e[i][j+1].se,e[i][j].se,0);            S1.addedge(e[i][j].se,e[i][j+1].se,e[i][j+1].fi-e[i][j].fi);                    }    }    S1.dijkstra(e[1][0].se);    ll ans=INF;    Rep(j,e[n].size()) {        int id=e[n][j].se;        gmin(ans,(ll)S1.d[id])    }    ans+=e[1][0].fi;    cout<<ans<<endl;        return 0;}