Luogu P3106 [USACO14OPEN]GPS的决斗Dueling GPS's(最短路)

P3106 [USACO14OPEN]GPS的决斗Dueling GPS's

题意

题目描述

Farmer John has recently purchased a new car online, but in his haste he accidentally clicked the "Submit" button twice when selecting extra features for the car, and as a result the car ended up equipped with two GPS navigation systems! Even worse, the two systems often make conflicting decisions about the route that FJ should take.

The map of the region in which FJ lives consists of \(N\) intersections ($2 \leq N \leq 10,000$) and \(M\) directional roads ($1 \leq M \leq 50,000$). Road \(i\) connects intersections \(A_i(1 \leq A_i \leq N)\) and \(B_i(1 \leq B_i \leq N)\). Multiple roads could connect the same pair of intersections, and a bi-directional road (one permitting two-way travel) is represented by two separate directional roads in opposite orientations. FJ's house is located at intersection $1$, and his farm is located at intersection \(N\). It is possible to reach the farm from his house by traveling along a series of directional roads.

Both GPS units are using the same underlying map as described above; however, they have different notions for the travel time along each road. Road \(i\) takes \(P_i\) units of time to traverse according to the first GPS unit, and \(Q_i\) units of time to traverse according to the second unit (each travel time is an integer in the range $1 \cdots 100,000$).

FJ wants to travel from his house to the farm. However, each GPS unit complains loudly any time FJ follows a road (say, from intersection \(X\) to intersection \(Y\)) that the GPS unit believes not to be part of a shortest route from X to the farm (it is even possible that both GPS units can complain, if FJ takes a road that neither unit likes).

Please help FJ determine the minimum possible number of total complaints he can receive if he chooses his route appropriately. If both GPS units complain when FJ follows a road, this counts as \(+2\) towards the total.

给你一个$N$个点的有向图,可能有重边.

有两个$GPS$定位系统,分别认为经过边$i$的时间为$P_i$和$Q_i$.

每走一条边的时候,如果一个系统认为走的这条边不是它认为的最短路,就会受到警告一次$T$

两个系统是分开警告的,就是说当走的这条边都不在两个系统认为的最短路范围内,就会受到$2$次警告.

如果边$(u,v)$不在$u$到$n$的最短路径上，这条边就受到一次警告，求从$1$到$n$最少受到多少次警告。

输入输出格式

输入格式：

Line $1$: The integers \(N\) and \(M\).

Line \(i\) describes road \(i\) with four integers: \(A_i \ B_i \ P_i \ Q_i\).

输出格式：

Line $1$: The minimum total number of complaints FJ can receive if he routes himself from his house to the farm optimally.

输入输出样例

输入样例：

5 7
3 4 7 1
1 3 2 20
1 4 17 18
4 5 25 3
1 2 10 1
3 5 4 14
2 4 6 5

输出样例：

1

说明

There are $5$ intersections and $7$ directional roads. The first road connects from intersection $3$ to intersection $4$; the first GPS thinks this road takes $7$ units of time to traverse, and the second GPS thinks it takes $1$ unit of time, etc.

If FJ follows the path $1$ → $2$ → $4$ → $5$, then the first GPS complains on the $1$ → $2$ road (it would prefer the $1$ → $3$ road instead). However, for the rest of the route $2$ → $4$ → $5$, both GPSs are happy, since this is a shortest route from $2$ to $5$ according to each GPS.

思路

太fAKe了。 --Mercury

我们发现，无论走到哪个点，$GPS$的警告都是使用到终点$n$的最短路径来判断的，所以我们先预处理出两台$GPS$到终点$n$的最短路长度$dis1,dis2$，这可以用反向跑最短路来实现。然后对于一条边$(u,v)$，如果$dis1[u]+len1(u,v)==dis1[v]$，那么第一台$GPS$是不会警告的；同样，如果$dis2[u]+len2(u,v)==dis2[v]$，那么第二台$GPS$是不会警告的。那么我们以此来改变每条边的边权为$GPS$警告的次数，然后再跑一遍最短路，就可以得到答案了。

AC代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL MAXN=1e4+5;
const LL MAXM=5e4+5;
LL n,m,dis[MAXN],dis1[MAXN],dis2[MAXN];
LL cnt,top[MAXN],to[MAXM],len[MAXM],nex[MAXM];
LL cnt1,top1[MAXN],to1[MAXM],len1[MAXM],nex1[MAXM];
LL cnt2,top2[MAXN],to2[MAXM],len2[MAXM],nex2[MAXM];
bool vis[MAXN];
inline LL read()
{
    LL re=0;char ch=getchar();
    while(!isdigit(ch)) ch=getchar();
    while(isdigit(ch)) re=(re<<3)+(re<<1)+ch-'0',ch=getchar();
    return re;
}
inline void add_edge(LL x,LL y,LL z){to[++cnt]=y,len[cnt]=z,nex[cnt]=top[x],top[x]=cnt;}
inline void add_edge1(LL x,LL y,LL z){to1[++cnt1]=y,len1[cnt1]=z,nex1[cnt1]=top1[x],top1[x]=cnt1;}
inline void add_edge2(LL x,LL y,LL z){to2[++cnt2]=y,len2[cnt2]=z,nex2[cnt2]=top2[x],top2[x]=cnt2;}
void SPFA()
{
    memset(dis,0x3f,sizeof dis);
    dis[1]=0;
    queue<LL>Q;
    Q.push(1);
    while(!Q.empty())
    {
        LL now=Q.front();Q.pop();
        vis[now]=false;
        for(int i=top[now];i;i=nex[i])
            if(dis[to[i]]>dis[now]+len[i])
            {
                dis[to[i]]=dis[now]+len[i];
                if(!vis[to[i]])
                {
                    vis[to[i]]=true;
                    Q.push(to[i]);
                }
            }
    }
}
void SPFA1()
{
    memset(dis1,0x3f,sizeof dis1);
    dis1[n]=0;
    queue<LL>Q;
    Q.push(n);
    while(!Q.empty())
    {
        LL now=Q.front();Q.pop();
        vis[now]=false;
        for(int i=top1[now];i;i=nex1[i])
            if(dis1[to1[i]]>dis1[now]+len1[i])
            {
                dis1[to1[i]]=dis1[now]+len1[i];
                if(!vis[to1[i]])
                {
                    vis[to1[i]]=true;
                    Q.push(to1[i]);
                }
            }
    }
}
void SPFA2()
{
    memset(dis2,0x3f,sizeof dis2);
    dis2[n]=0;
    queue<LL>Q;
    Q.push(n);
    while(!Q.empty())
    {
        LL now=Q.front();Q.pop();
        vis[now]=false;
        for(int i=top2[now];i;i=nex2[i])
            if(dis2[to2[i]]>dis2[now]+len2[i])
            {
                dis2[to2[i]]=dis2[now]+len2[i];
                if(!vis[to2[i]])
                {
                    vis[to2[i]]=true;
                    Q.push(to2[i]);
                }
            }
    }
}
int main()
{
    n=read(),m=read();
    while(m--)
    {
        int x=read(),y=read(),z1=read(),z2=read();
        add_edge1(y,x,z1);
        add_edge2(y,x,z2);
    }
    SPFA1(),SPFA2();
    for(int i=1;i<=n;i++)
        for(int j=top1[i];j;j=nex1[j])
        {
            int l=2;
            if(dis1[to1[j]]==dis1[i]+len1[j]) l--;
            if(dis2[to2[j]]==dis2[i]+len2[j]) l--;
            add_edge(to1[j],i,l);
        }
    SPFA();
    printf("%lld",dis[n]);
    return 0;
}