(1): Shortest Path

Others 2019-08-25 11:46:44 views: null

Dijkstra's algorithm

1003 Emergency (25 分)
 

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input Specification:

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (≤) - the number of cities (and the cities are numbered from 0 to N−1), M - the number of roads, C​1​​ and C​2​​ - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c​1​​, c​2​​ and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C​1​​ to C​2​​.

Output Specification:

For each test case, print in one line two numbers: the number of different shortest paths between C​1​​ and C​2​​, and the maximum amount of rescue teams you can possibly gather. All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input:

5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

Sample Output:

2 4

#include <cstdio>
#include <algorithm>

using namespace std;
const int MAXN=510;
const int INF=1000000000;
int n,m,c1,c2;
int g[MAXN][MAXN];
bool vis[MAXN]={false};
int d[MAXN];
int weight[MAXN];
int w[MAXN]={0},num[MAXN]={0};

void dijk(int s)
{
    fill(d,d+MAXN,INF);
    d[s]=0;num[s]=1;w[s]=weight[s];
    
    for(int i=0;i<n;i++)
    {
        int u=-1,minl=INF;
        for(int j=0;j<n;j++)
        {
            if(vis[j]==false&&d[j]<minl)
            {
                u=j;
                minl=d[j];
            }
        }
        if(u==-1) return;
        vis[u]=true;
        for(int j=0;j<n;j++)
        {
            if(vis[j]==false&&g[u][j]!=INF)
            {
                if(d[u]+g[u][j]<d[j])
                {
                    d[j]=d[u]+g[u][j];
                    w[j]=w[u]+weight[j];
                    num[j]=num[u];
                        
                }else if(d[u]+g[u][j]==d[j])
                {
                    num[j]+=num[u];
                    if(w[u]+weight[j]>w[j]) w[j]=w[u]+weight[j];
                }    
            }    
        }         
    }    
}

int main()
{
    scanf("%d%d%d%d",&n,&m,&c1,&c2);
    fill(g[0],g[0]+MAXN*MAXN,INF);
    for(int i=0;i<n;i++) scanf("%d",&weight[i]);
    int x,y,l;
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d",&x,&y,&l);
        g[x][y]=g[y][x]=l;
    }
    dijk(c1);
    printf("%d %d",num[c2],w[c2]);
    return 0;
}

 Seeking a shortest length, add, modify in place to update the shortest distance: the right point, while the right number.

 

1030 Travel Plan (30 分)
 

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (≤) is the number of cities (and hence the cities are numbered from 0 to N−1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:

City1 City2 Distance Cost

where the numbers are all integers no more than 500, and are separated by a space.

Output Specification:

For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.

Sample Input:

4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20

Sample Output:

0 2 3 3 40
#include <cstdio>
#include <algorithm>
#include <vector>

using namespace std;
const int MAXN=510;
const int INF=1000000000;
int n,m,st,en;
int g[MAXN][MAXN];
int d[MAXN];
bool vis[MAXN]={false};
vector<int> pre[MAXN];
void dijk(int s)
{
    fill(d,d+MAXN,INF);
    d[s]=0;
    for(int i=0;i<n;i++)
    {
        int u=-1,minl=INF;
        for(int j=0;j<n;j++)
        {
            if(vis[j]==false&&d[j]<minl)
            {
                u=j;
                minl=d[j];
            }
        }
        if(u==-1) return;
        vis[u]=true;
        for(int v=0;v<n;v++)
        {
            if(vis[v]==false&&g[u][v]!=INF)
            {
                if(d[v]>d[u]+g[u][v])
                {
                    d[v]=d[u]+g[u][v];
                    pre[v].clear();
                    pre[v].push_back(u);
                }else if(d[v]==d[u]+g[u][v])
                {
                    pre[v].push_back(u);
                }
            }
        }
    }
}
int mincost=INF,c[MAXN][MAXN];
vector<int> temp,ans;
void dfs(int v)
{
    if(v==st)
    {
        temp.push_back(v);
        int sumcost=0;
        for(int i=temp.size()-1;i>0;i--)
        {
            sumcost+=c[temp[i]][temp[i-1]];
        }
        if(sumcost<mincost)
        {
            mincost=sumcost;
            ans=temp;
        }
        temp.pop_back();
        return;
    }
    temp.push_back(v);
    for(int i=0;i<pre[v].size();i++)
    {
        dfs(pre[v][i]);
    }
    temp.pop_back();
    
}
int main()
{
    int x,y,dis,cos;
    fill(g[0],g[0]+MAXN*MAXN,INF);
    fill(c[0],c[0]+MAXN*MAXN,INF);
    scanf("%d%d%d%d",&n,&m,&st,&en);
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d%d",&x,&y,&dis,&cos);
        g[x][y]=g[y][x]=dis;
        c[x][y]=c[y][x]=cos;
    }
    dijk(st);
    dfs(en);
    for(int i=ans.size()-1;i>=0;i--)
    {
        printf("%d ",ans[i]);
    }
    printf("%d %d"d [in] mincost);
    return  0 ; 
}

Due to the shortest distance, determines the minimum price the only direct Dijkstra + arrays to store precursors point is the optimal solution. It may also be a more general method, to obtain the shortest distance, and able to save a point to each of the shortest path with the vector, may not be unique because it is a multi-dimensional vector. DFS then find the least price. If no unique best, it can also be stored vectors.

The main thing is routine.

 

1018 Public Bike Management (30 分)
 

There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city.

The Public Bike Management Center (PBMC) keeps monitoring the real-time capacity of all the stations. A station is said to be in perfect condition if it is exactly half-full. If a station is full or empty, PBMC will collect or send bikes to adjust the condition of that station to perfect. And more, all the stations on the way will be adjusted as well.

When a problem station is reported, PBMC will always choose the shortest path to reach that station. If there are more than one shortest path, the one that requires the least number of bikes sent from PBMC will be chosen.

The above figure illustrates an example. The stations are represented by vertices and the roads correspond to the edges. The number on an edge is the time taken to reach one end station from another. The number written inside a vertex S is the current number of bikes stored at S. Given that the maximum capacity of each station is 10. To solve the problem at S​3​​, we have 2 different shortest paths:

  1. PBMC -> S​1​​ -> S​3​​. In this case, 4 bikes must be sent from PBMC, because we can collect 1 bike from S​1​​ and then take 5 bikes to S​3​​, so that both stations will be in perfect conditions.

  2. PBMC -> S​2​​ -> S​3​​. This path requires the same time as path 1, but only 3 bikes sent from PBMC and hence is the one that will be chosen.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 numbers: C​max​​ (≤), always an even number, is the maximum capacity of each station; N (≤), the total number of stations; S​p​​, the index of the problem station (the stations are numbered from 1 to N, and PBMC is represented by the vertex 0); and M, the number of roads. The second line contains N non-negative numbers C​i​​ (,) where each C​i​​ is the current number of bikes at S​i​​ respectively. Then M lines follow, each contains 3 numbers: S​i​​, S​j​​, and T​ij​​ which describe the time T​ij​​ taken to move betwen stations S​i​​ and S​j​​. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print your results in one line. First output the number of bikes that PBMC must send. Then after one space, output the path in the format: 0. Finally after another space, output the number of bikes that we must take back to PBMC after the condition of S​p​​ is adjusted to perfect.

Note that if such a path is not unique, output the one that requires minimum number of bikes that we must take back to PBMC. The judge's data guarantee that such a path is unique.

Sample Input:

10 3 3 5
6 7 0
0 1 1
0 2 1
0 3 3
1 3 1
2 3 1

Sample Output:

3 0->2->3 0

 

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int MAXN=510;
const int INF=100000000;
int cm,n,sp,m;
int c[MAXN];
int g[MAXN][MAXN]; int d[MAXN]; bool vis[MAXN]={false};
vector <int> pre[MAXN];
void dijk(int s)
{
    fill(d,d+MAXN,INF);
    d[s]=0;
    for(int i=0;i<=n;i++)
    {
        int u=-1,minl=INF;
        for(int j=0;j<=n;j++)
        {
            if(vis[j]==false&&d[j]<minl)
            {
                u=j;minl=d[j];
            }
        }
        if(u==-1) return;
        vis[u]=true;
        for(int v=0;v<=n;v++)
        {
            if(vis[v]==false&&g[u][v]!=INF)
            {
                if(d[u]+g[u][v]<d[v])
                {
                    d[v]=d[u]+g[u][v];
                    pre[v].clear();
                    pre[v].push_back(u);
                }else if(d[v]==d[u]+g[u][v])
                {
                    pre[v].push_back(u);
                }
            }
        }
    }
}
vector<int> temp,ans;
int minneed=INF,minremain=INF;
void dfs(int v)
{
    if(v==0)
    {
        int need=0,remain=0;
        for(int i=temp.size()-1;i>=0;i--)
        {
            if(c[temp[i]]>0) remain+=c[temp[i]];
            else
            {
                if(remain>abs(c[temp[i]])) remain+=c[temp[i]];
                else
                {
                    need+=abs(c[temp[i]])-remain;
                    remain=0;
                }
            }
        }
        
        if(need<minneed)
        {
            ans=temp;
            minneed=need;
            minremain=remain;
        }else if(need==minneed&&remain<minremain)
        {
            minremain=remain;
            ans=temp;
        }
        return;
    }
    temp.push_back(v);
    for(int i=0;i<pre[v].size();i++)
    {
        dfs(pre[v][i]);
    }
    temp.pop_back();
}




int main()
{
    scanf("%d%d%d%d",&cm,&n,&sp,&m);
    c[0]=0;cm/=2;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&c[i]);c[i]-=cm;
    }
    fill(g[0],g[0]+MAXN*MAXN,INF);
    int x,y,z;
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d",&x,&y,&z);
        g[y][x]=g[x][y]=z;
    }
    dijk(0);
    dfs(sp);    
    
    printf("%d ",-minneed);
    printf("0");
    for(int i=ans.size()-1;i>=0;i--)
    {
        printf("->%d",ans[i]);
    }
    printf(" %d",minremain);
    
    return 0;
}

Also part Dijkstra + DFS, the routine is not to say, initially thought can be determined by the need to fill all still need to get the right point of the path and statistically, in the circumstances, considering only from the number of conservation is flawed because you must go to all the points in a trip back to normal, then the rest of the band came back again. If it is a return to repair all the points, you should be able to count the total number of directly. This is in fact the case with the previous minimum, remaining at the destination to the least, two of the selection criteria.

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Origin www.cnblogs.com/fremontxutheultimate/p/11364668.html
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