• 分支限界法 求电路布线的最短路径


    //用队列进行广度优先搜索


    #include<iostream> #include<queue> #include<fstream> using namespace std; //import the grid data; ifstream fin("map.txt");//这个map是7行7列,请以这个为例子来理解这个程序 typedef struct Position { int row; int col; } Posi; //find the shortest path for the grid bool FindPath(Posi start,Posi finish,int & PathLen,int **&grid,Posi *& path,int n,int m) { //if the start position is the finish position if((start.row == finish.row) && (start.col == finish.col)) { PathLen = 0; return true; } Position offset[4]; offset[0].row = -1;//up offset[0].col = 0; offset[1].row = 1;//down offset[1].col = 0; offset[2].row = 0;//left offset[2].col = -1; offset[3].row = 0;//right offset[3].col = 1; Posi here,nbr; here.row = start.row; here.col = start.col; int NumOfNbrs = 4;//ajacent position; grid[start.row][start.col] = 2;//init the start position's length with value 2, queue<Posi> Q; do { for(int firdex = 0;firdex < NumOfNbrs;firdex++) { nbr.row = here.row + offset[firdex].row; nbr.col = here.col + offset[firdex].col; if(grid[nbr.row][nbr.col] == 0)//this position haven't been visted { grid[nbr.row][nbr.col] = grid[here.row][here.col] + 1; if((nbr.row == finish.row) && (nbr.col == finish.col))//find the shortest path { break; } Q.push(nbr); } } if((nbr.row == finish.row) && (nbr.col ==finish.col)) { break;//wiring was completed } if(Q.empty())//if queue is empty { return false;//no result } here = Q.front(); Q.pop(); }while(true); //traceback the shortest path PathLen = grid[finish.row][finish.col]-2; path = new Posi[PathLen]; here = finish; for(int firdex = PathLen-1;firdex >=0;firdex--) { path[firdex] = here; for(int secdex = 0;secdex < NumOfNbrs;secdex++) { nbr.row = here.row + offset[secdex].row; nbr.col = here.col + offset[secdex].col; if(grid[nbr.row][nbr.col] == firdex+2)//It is the nbr's grid that why the grid[nbr.row][nbr.col] can give index of "firdex+2" { break; } } here =nbr;//move to the previous node } return true; } //allocate memory for the grid void InitGrid(int **&grid,int n,int m) { grid = new int*[n+2]; for(int firdex = 0;firdex < n+2;firdex++) grid[firdex] = new int[m+2]; //set the bound for(int index = 0;index < m+2;index++)//top an bottom { grid[0][index] = grid[n+1][index] =1; } for(int index = 0;index < n+2;index++) { grid[index][0] = grid[index][m+1] = 1; } for(int firdex = 1;firdex < n+1;firdex++) { for(int secdex = 1;secdex < m+1;secdex++) fin>>grid[firdex][secdex]; } } //destroy the resource for the grid void Destroy(int ** &grid,int n,int m) { if(grid != NULL) { for(int firdex = 0;firdex < n+2;firdex++) { delete [] grid[firdex]; grid[firdex] = NULL; } delete grid; grid = NULL; } } int main(void) { int m = 0,n = 0; Posi start,finish; int PathLength = 0; Posi * path = NULL; int ** grid = NULL; cout<<"Please input the m and n of the grid:"<<endl; cin>>n>>m; cout<<"Please input the start position:"<<endl; cout<<"start:row ="; cin>>start.row; cout<<"start:col ="; cin>>start.col; cout<<"Please input the finish position:"<<endl; cout<<"finish:row ="; cin>>finish.row; cout<<"finish:col ="; cin>>finish.col; InitGrid(grid,n,m); cout<<"the map resource:"<<endl; for(int firdex = 1;firdex < n+1;firdex++) { for(int secdex = 1;secdex < m+1;secdex++) cout<<grid[firdex][secdex]<<" "; cout<<endl; } cout<<endl; FindPath(start,finish,PathLength,grid,path,n,m); cout<<"The shortest path of wiring is :"<<PathLength<<endl; cout<<"The path if follow:"<<endl; for(int index = 0;index < PathLength;index++) { cout<<"("<<path[index].row<<","<<path[index].col<<")"; if(index < PathLength-1) cout<<"-->"; } cout<<endl; //Destory the resource of grid Destroy(grid,n,m); //release the path's resource if(path != NULL) { delete [] path; path = NULL; } return 0; }
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  • 原文地址:https://www.cnblogs.com/Linkliu/p/wiringProblem.html
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