4289: PA2012 Tax

最新推荐文章于 2018-08-28 11:54:39 发布

最新推荐文章于 2018-08-28 11:54:39 发布 · 452 阅读

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4289: PA2012 Tax

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 271 Solved: 90
[ Submit][ Status][ Discuss]

Description

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

Input

Output

Sample Input

4 5
1 2 5
1 3 2
2 3 1
2 4 4
3 4 8

Sample Output

HINT

Source

[ Submit][ Status][ Discuss]

一个点的点权是由出入边共同决定，不好统计

先将原图转化，点->边，边->点，每条边又拆成两个点

对于每个点，将从它连出去的所有出边，按照权值升序排好，做差分

相邻边连边，权值小到大的边权为差分值，从大到小的，边权为0

对于每条边拆分的两个点，分别连权值为边权的边

考虑在新图中跑最短路，，从某个出边代表的点，转移到它的反向边

通过差分的权值进行方向调整，于是这样就能跑出原图的最短路了

画个图比较好理解，最后起点和终点的边权注意特判

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<ext/pb_ds/priority_queue.hpp>
using namespace std;
 
const int maxn = 1E5 + 10;
const int maxm = 2E5 + 20;
const int N = maxm * 2;
typedef long long LL;
const LL INF = 1E16;
 
struct E{
    int to,w; E(){}
    E(int to,int w): to(to),w(w){}
};
 
struct data{
    int Num; LL d; data(){}
    data(int Num,LL d): Num(Num),d(d){}
    bool operator < (const data &B) const {return d > B.d;}
};
typedef __gnu_pbds::priority_queue<data,less<data>,__gnu_pbds::pairing_heap_tag> Heap;
 
int n,m,tot = 2,x[maxm],y[maxm],w[maxm],A[maxm],B[maxm];
LL dis[N]; bool inq[N];
 
vector <E> v[N];
vector <int> G[maxn];
Heap Q; Heap::point_iterator id[N];
 
bool cmp(const int &a,const int &b) {return w[a] < w[b];}
int getint()
{
    char ch = getchar(); int ret = 0;
    while (ch < '0' || '9' < ch) ch = getchar();
    while ('0' <= ch && ch <= '9')
        ret = ret*10 + ch - '0',ch = getchar();
    return ret;
}
 
int main()
{
    #ifdef DMC
        freopen("DMC.txt","r",stdin);
    #endif
 
    n = getint(); m = getint();
    for (int i = 1; i <= m; i++)
    {
        A[i] = tot++; B[i] = tot++;
        x[i] = getint(); y[i] = getint(); w[i] = getint();
        G[x[i]].push_back(i); G[y[i]].push_back(i);
        v[A[i]].push_back(E(B[i],w[i]));
        v[B[i]].push_back(E(A[i],w[i]));
    }
    for (int i = 2; i < n; i++)
    {
        if (!G[i].size()) continue;
        sort(G[i].begin(),G[i].end(),cmp);
        int last = x[G[i][0]] == i ? A[G[i][0]] : B[G[i][0]];
        for (int j = 1; j < G[i].size(); j++)
        {
            int now = x[G[i][j]] == i ? A[G[i][j]] : B[G[i][j]];
            v[last].push_back(E(now,w[G[i][j]] - w[G[i][j-1]]));
            v[now].push_back(E(last,0)); last = now;
        }
    }
    for (int i = 0; i < G[1].size(); i++)
    {
        int k = x[G[1][i]] == 1 ? A[G[1][i]] : B[G[1][i]];
        v[1].push_back(E(k,w[G[1][i]]));
    }
    for (int i = 0; i < G[n].size(); i++)
    {
        int k = y[G[n][i]] == n ? A[G[n][i]] : B[G[n][i]];
        v[k].push_back(E(tot,w[G[n][i]]));
    }
     
    id[1] = Q.push(data(1,0)); inq[1] = 1;
    for (int i = 2; i <= tot; i++) dis[i] = INF;
    while (!Q.empty())
    {
        int k = Q.top().Num; Q.pop();
        for (int i = 0; i < v[k].size(); i++)
        {
            E e = v[k][i];
            if (dis[e.to] > dis[k] + 1LL * e.w)
            {
                dis[e.to] = dis[k] + 1LL * e.w;
                if (!inq[e.to]) inq[e.to] = 1,id[e.to] = Q.push(data(e.to,dis[e.to]));
                else Q.modify(id[e.to],data(e.to,dis[e.to]));
            }
        }
    }
    cout << dis[tot] << endl;
    return 0;
}