BZOJ 3445: [Usaco2014 Feb] Roadblock

Description

一个图, \(n\) 个点 \(m\) 条边,求将一条边距离翻倍后使 \(1-n\) 最短路径增加的最大增量.

Sol

Dijstra.

先跑一边最短路,然后枚举最短路,将路径翻倍然后跑Dijstra...

因为不在最短路径上的边没用贡献,然后最短路径最长为 \(n-1\)

复杂度 \(O(nmlogm\)

Code

/**************************************************************
    Problem: 3445
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:32 ms
    Memory:3332 kb
****************************************************************/

#include <cstdio>
#include <cstring>
#include <utility>
#include <vector>
#include <queue>
#include <functional>
#include <iostream>
using namespace std;

typedef long long LL;
typedef pair< LL,int > pr;
#define mpr make_pair
const int N = 255;

int n,m,cnt;
int b[N],p[N];
LL ans,disn;
LL d[N];
struct Edge{ int fr,to;LL w; }edge[N*N*2];
vector< int > g[N];
priority_queue< pr,vector< pr >,greater< pr > > q;
vector< int > path;

inline int in(int x=0){ scanf("%d",&x);return x; }
void Add_Edge(int fr,int to,int w){
    edge[++cnt]=(Edge){ fr,to,w };
    g[fr].push_back(cnt);
}
void GetPath(){
    for(int x=n;x!=1;x=edge[p[x]].fr) path.push_back(p[x]);
}
void Dijstra(int s,int fst){
    memset(b,0,sizeof(b));
    memset(d,0x3f,sizeof(d));
    d[s]=0,q.push(mpr(0LL,s));
    for(int x;!q.empty();){
        x=q.top().second,q.pop();if(b[x]) continue;b[x]=1;
        for(int i=0,lim=g[x].size();i<lim;i++){
            int v=edge[g[x][i]].to;LL w=edge[g[x][i]].w;
//          cout<<x<<" "<<v<<" "<<w<<" "<<d[x]<<" "<<d[v]<<endl;
            if(d[x]+w < d[v]){
                d[v]=d[x]+w,p[v]=g[x][i];
                q.push(mpr(d[v],v));
            }
        }
    }
//  cout<<d[n]<<endl;
    if(fst) disn=d[n],GetPath();
    else ans=max(ans,d[n]-disn);
}
int main(){
    n=in(),m=in();
    for(int i=1,u,v,w;i<=m;i++) u=in(),v=in(),w=in(),Add_Edge(u,v,w),Add_Edge(v,u,w);
    Dijstra(1,1);

//  for(int i=0,lim=path.size();i<lim;i++) cout<<edge[path[i]].fr<<" "<<edge[path[i]].to<<" "<<edge[path[i]].w<<endl;

    for(int i=0,lim=path.size();i<lim;i++) edge[path[i]].w*=2,Dijstra(1,0),edge[path[i]].w/=2;
    cout<<ans<<endl;
    return 0;
}