第十六周:( Sicily1321) Robot(c++)

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原题链接:
http://soj.sysu.edu.cn/1321

思路:这是一道求单源最短路的题,笔者分别用Dijkstra和Bellman-Ford两种算法求解。我们知道,Dijkstra算法对只含有正权边的图管用,一般情况下(不用二分堆)复杂度是O(V^2)。而BellmanFord算法对含有负权边的图一样管用,其复杂度为O(V*E),但如果图是稀疏图,此算法的效率更好!以下代码,Bellman-Ford算法的ac时间为0.03s,Dijkstra的ac时间为0.37s,这应该也是因为图比较稀疏的原因。

代码一(Dijkstra算法):

#include <iostream>using namespace std;#define INF 999999999int Matrix[105][105],DisMatrix[105][105];bool Visited[105][105];int Dijkstra(pair<int,int> start,pair<int,int> end, int m, int n){    pair<int,int> MinDistancePoint;    int nPoint=m*n;    //初始化    Visited[start.first][start.second]=true;    DisMatrix[start.first][start.second]=Matrix[start.first][start.second];    MinDistancePoint=start;    for(int i=1;i<nPoint;i++){        int x=MinDistancePoint.first,y=MinDistancePoint.second;        //更新到start点的值        if((x-1>=1)&&(!Visited[x-1][y])&&((DisMatrix[x][y]+Matrix[x-1][y])<DisMatrix[x-1][y]))            DisMatrix[x-1][y]=DisMatrix[x][y]+Matrix[x-1][y];        if((y-1>=1)&&(!Visited[x][y-1])&&((DisMatrix[x][y]+Matrix[x][y-1])<DisMatrix[x][y-1]))            DisMatrix[x][y-1]=DisMatrix[x][y]+Matrix[x][y-1];        if((x+1<=m)&&(!Visited[x+1][y])&&((DisMatrix[x][y]+Matrix[x+1][y])<DisMatrix[x+1][y]))            DisMatrix[x+1][y]=DisMatrix[x][y]+Matrix[x+1][y];        if((y+1<=n)&&(!Visited[x][y+1])&&((DisMatrix[x][y]+Matrix[x][y+1])<DisMatrix[x][y+1]))            DisMatrix[x][y+1]=DisMatrix[x][y]+Matrix[x][y+1];        int MinDistance=INF;        //查找下一个visited的点(即未visited中最近的点)        for(int a=1;a<=m;a++)            for(int b=1;b<=n;b++){                if((!Visited[a][b])&&(DisMatrix[a][b]<MinDistance)){                    MinDistance=DisMatrix[a][b];                    MinDistancePoint.first=a;                    MinDistancePoint.second=b;                }            }        //访问最小的点        Visited[MinDistancePoint.first][MinDistancePoint.second]=true;    }    return DisMatrix[end.first][end.second];}int main(){    int T;    cin>>T;    while(T--){        int m,n;        cin>>m>>n;        if((m==0)||(n==0))            cout<<0<<endl;        for(int i=1;i<=m;i++)            for(int j=1;j<=n;j++){                cin>>Matrix[i][j];                DisMatrix[i][j]=INF;                Visited[i][j]=false;            }        pair<int,int> start,end;        cin>>start.first>>start.second>>end.first>>end.second;        cout<<Dijkstra(start,end,m,n)<<endl;    }    return 0;}

代码二(Bellman-Ford算法):

#include <iostream>#include <queue>using namespace std;#define INF 999999999int Matrix[105][105],DisMatrix[105][105];int BellmanFord(pair<int,int> start,pair<int,int> end, int m, int n){    queue <pair<int,int>> UpdatePoint;    int nPoint=m*n;    //初始化    DisMatrix[start.first][start.second]=Matrix[start.first][start.second];    UpdatePoint.push(start);    for(int i=1;i<nPoint;i++){        queue <pair<int,int>> NextUpdatePoint;        //更新        while(!UpdatePoint.empty()){            pair<int,int> point;            point=UpdatePoint.front();            int x=point.first;            int y=point.second;            if((x-1>=1)&&((DisMatrix[x][y]+Matrix[x-1][y])<DisMatrix[x-1][y])){                DisMatrix[x-1][y]=DisMatrix[x][y]+Matrix[x-1][y];                NextUpdatePoint.push(make_pair(x-1,y));            }            if((y-1>=1)&&((DisMatrix[x][y]+Matrix[x][y-1])<DisMatrix[x][y-1])){                DisMatrix[x][y-1]=DisMatrix[x][y]+Matrix[x][y-1];                NextUpdatePoint.push(make_pair(x,y-1));            }            if((x+1<=m)&&((DisMatrix[x][y]+Matrix[x+1][y])<DisMatrix[x+1][y])){                DisMatrix[x+1][y]=DisMatrix[x][y]+Matrix[x+1][y];                NextUpdatePoint.push(make_pair(x+1,y));            }            if((y+1<=n)&&((DisMatrix[x][y]+Matrix[x][y+1])<DisMatrix[x][y+1])){                DisMatrix[x][y+1]=DisMatrix[x][y]+Matrix[x][y+1];                NextUpdatePoint.push(make_pair(x,y+1));            }            UpdatePoint.pop();        }        //更新下轮需要更新的点        while(!NextUpdatePoint.empty()){            UpdatePoint.push(NextUpdatePoint.front());            NextUpdatePoint.pop();        }    }    return DisMatrix[end.first][end.second];}int main(){    int T;    cin>>T;    while(T--){        int m,n;        cin>>m>>n;        if((m==0)||(n==0))            cout<<0<<endl;        for(int i=1;i<=m;i++)            for(int j=1;j<=n;j++){                cin>>Matrix[i][j];                DisMatrix[i][j]=INF;            }        pair<int,int> start,end;        cin>>start.first>>start.second>>end.first>>end.second;        cout<<BellmanFord(start,end,m,n)<<endl;    }    return 0;}
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