俄罗斯套娃奖品

中兴捧月程序大赛初赛中有一道关于俄罗斯套娃奖品的题目，具体内容如下：

 

伊万洛夫在比武大会上力克群雄，成为新一届“草原雄鹰”，为部落赢得了莫大荣誉。首领决定要重重奖赏，他对伊万洛夫说：“孩子，你是知道的，面前的这片草原，南北向和东西向的道路纵横交错。现在，路口放着纯金打造的俄罗斯娃娃，重量大小不等，重的都能装下轻的。你可以沿着道路飞奔，拾取路口的娃娃，要求是任何时刻必须是一个套娃，装好后就不能再拆开了。注意不要走重复路。”

请你为伊万洛夫规划路线，使得他能够有最大的收获。
Input:  cross.txt
        输入包括多组测试用例；
        每个测试用例开始是一对整数<R, C>，R表示东西向道路数，C表示南北向道路总数；接下来R行，每行包括C个正整数W[r,c] ，分别表示第r条东西向道路与第c条南北向道路交叉处路口放置的俄罗斯娃娃的重量。
Output:
        输出能有最大收获的路径规划。
假设1：
   cross.txt
   2     7
   1     2   13   6   7  12  11
   14   3   4     5   8   9   10
输出：
1 2 3 4 5 6 7 8 9 10 11 12

假设2：
   cross.txt
   5    5
   1   16   15   14   13
   2   17   24   23   12
   3   18   25   22   11
   4   19   20   21   10
   5    6     7     8     9
输出：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

注释:
1)从<0,0>出发;
2)路线不能重复;
3)不要求最后回到出发点。

需要注意，路口处放置的套娃重量可以为0，经过的路口不能过儿不拾。

首先考虑，如果套娃的重量不可以为0的话，算法相对简单，一种实现方法如下（使用类似于迷宫问题的回溯法可以解决这个问题）：

#include <iostream>#include <vector>#include <stack>#include <string>#include <fstream>#include <sstream>#include <iomanip>using namespace std;typedef struct _position{int x,y;}POSITION;stack<POSITION> position_stack,record_stack;POSITION psn;int nR=0,nC=0;size_t sum=0,sumres=0;void visit(int **map,int x,int y);int main(){// open fileifstream infile("cross.txt");if (!infile){cerr<<"error, open cross.txt"<<endl;cin.get();return -1;}vector<string> sLineVec;string strLine,strWord;bool bFirstLine=true;while (getline(infile,strLine)){if (bFirstLine)// read R and C{istringstream iss(strLine);int i=1;while (iss>>strWord){if (i==1){nR=atoi(strWord.c_str());// R}else{nC=atoi(strWord.c_str());// C}++i;}}else// read weight{sLineVec.push_back(strLine);}bFirstLine=false;}// new an array to stands for the map of weightint **map=new int*[nR];for(int i=0; i!=nR; ++i){map[i]=new int[nC];}// initialize the mapint i=0,j;for (vector<string>::const_iterator it=sLineVec.begin(); it!=sLineVec.end(); ++it,++i){j=0;istringstream iss(*it);string strWord;while (iss>>strWord){map[i][j++]=atoi(strWord.c_str());}}// show the mapfor (int i=0; i!=nR; ++i){for (int j=0; j!=nC; ++j){cout<<setw(2)<<map[i][j]<<" ";}cout<<endl;}// start form map[0][0]sum=map[0][0];psn.x=0;psn.y=0;position_stack.push(psn);visit(map,0,0);cout<<"The biggest weight is "<<sumres<<endl;// show the answerstack<POSITION> tmpStack;POSITION tmpPSN;while (record_stack.size()){tmpPSN=record_stack.top();tmpStack.push(tmpPSN);record_stack.pop();}while (tmpStack.size()){tmpPSN=tmpStack.top();cout<<map[tmpPSN.x][tmpPSN.y]<<" ";tmpStack.pop();}cout<<endl;// free the heapfor (int i=0; i!=nR; ++i){delete []map[i];}delete []map;cin.get();return 0;}void visit(int **map,int x,int y){if (x+1<nR && map[x][y]<map[x+1][y])// down{sum+=map[x+1][y];psn.x=x+1;psn.y=y;position_stack.push(psn);visit(map,x+1,y);}if (x-1>=0 && map[x][y]<map[x-1][y])// up{sum+=map[x-1][y];psn.x=x-1;psn.y=y;position_stack.push(psn);visit(map,x-1,y);}if (y+1<nC && map[x][y]<map[x][y+1])// right{sum+=map[x][y+1];psn.x=x;psn.y=y+1;position_stack.push(psn);visit(map,x,y+1);}if (y-1>=0 && map[x][y]<map[x][y-1])// left{sum+=map[x][y-1];psn.x=x;psn.y=y-1;position_stack.push(psn);visit(map,x,y-1);}// check and save if (sum>sumres){sumres=sum;record_stack=position_stack;}if (position_stack.size()){psn=position_stack.top();sum-=map[psn.x][psn.y];position_stack.pop();// synif (position_stack.size()==0)// have tried all{cout<<"Get the answer..."<,endl;return;}}}

如果可以允许套娃的重量为0，即允许某一路口处没有放置套娃，那么需要考虑的情况会多一些，一种实现方法如下：

/* * OS: Windows Server 2003 * Compiler: VS2008 * * These codes conside the case having zero weight*/#include <iostream>#include <vector>#include <stack>#include <string>#include <fstream>#include <sstream>#include <iomanip>using namespace std;typedef struct _position{int x,y;}POSITION;stack<POSITION> position_stack,record_stack,postmp_stack;POSITION psn;int nR=0,nC=0;size_t sum=0,sumres=0;void visit(int **map,int x,int y);int main(){// open fileifstream infile("cross.txt");if (!infile){cout<<"error, open cross.txt"<<endl;cin.get();return -1;}vector<string> sLineVec;string strLine,strWord;bool bFirstLine=true;while (getline(infile,strLine)){if (bFirstLine)// read R and C{istringstream iss(strLine);int i=1;while (iss>>strWord){if (i==1){nR=atoi(strWord.c_str());// R}else{nC=atoi(strWord.c_str());// C}++i;}}else// read weight{sLineVec.push_back(strLine);}bFirstLine=false;}// new an array to stands for the map of weightint **map=new int*[nR];for(int i=0; i!=nR; ++i){map[i]=new int[nC];}// initialize the mapint i=0,j;for (vector<string>::const_iterator it=sLineVec.begin(); it!=sLineVec.end(); ++it,++i){j=0;istringstream iss(*it);string strWord;while (iss>>strWord){map[i][j++]=atoi(strWord.c_str());}}// show the mapfor (int i=0; i!=nR; ++i){for (int j=0; j!=nC; ++j){cout<<setw(2)<<map[i][j]<<" ";}cout<<endl;}// start form map[0][0]sum=map[0][0];psn.x=0;psn.y=0;position_stack.push(psn);visit(map,0,0);cout<<"The biggest weight is: "<<sumres<<endl;cout<<"The perhaps path is: ";// show the answerstack<POSITION> tmpStack;POSITION tmpPSN;while (record_stack.size()){tmpPSN=record_stack.top();tmpStack.push(tmpPSN);record_stack.pop();}while (tmpStack.size()){tmpPSN=tmpStack.top();cout<<map[tmpPSN.x][tmpPSN.y]<<" ";tmpStack.pop();}cout<<endl;// free the heapfor (int i=0; i!=nR; ++i){delete []map[i];}delete []map;cin.get();return 0;}void visit(int **map,int x,int y){if (x+1<nR && map[x][y]<map[x+1][y] || x+1<nR && map[x+1][y]==0)// down{// checkbool bHave=false,bNotbigger=false;if (map[x][y]==-1 && map[x+1][y]!=0 || map[x+1][y]==0)// two cases need to check{postmp_stack=position_stack;POSITION psnnext;psnnext.x=x+1;psnnext.y=y;while (postmp_stack.size()){psn=postmp_stack.top();if (psnnext.x==psn.x && psnnext.y==psn.y)// have this point yet {bHave=true;break;}if (map[x+1][y]!=0){if (map[x+1][y]<=map[psn.x][psn.y])// not less than before{bNotbigger=true;break;}}postmp_stack.pop();}}// if ok, then save this pointif (!bHave && !bNotbigger){sum+=map[x+1][y];psn.x=x+1;psn.y=y;position_stack.push(psn);if (map[x+1][y]==0){map[x+1][y]=-1;}visit(map,x+1,y);}}if (x-1>=0 && map[x][y]<map[x-1][y] || x-1>=0 && map[x-1][y]==0)// up{// checkbool bHave=false,bNotbigger=false;if (map[x][y]==-1 && map[x-1][y]!=0 || map[x-1][y]==0)// two cases need to check{postmp_stack=position_stack;POSITION psnnext;psnnext.x=x-1;psnnext.y=y;while (postmp_stack.size()){psn=postmp_stack.top();if (psnnext.x==psn.x && psnnext.y==psn.y)// have this point yet {bHave=true;break;}if (map[x-1][y]!=0){if (map[x-1][y]<=map[psn.x][psn.y])// not less than before{bNotbigger=true;break;}}postmp_stack.pop();}}// if ok, then save this pointif (!bHave && !bNotbigger){sum+=map[x-1][y];psn.x=x-1;psn.y=y;position_stack.push(psn);if (map[x-1][y]==0){map[x-1][y]=-1;}visit(map,x-1,y);}}if (y+1<nC && map[x][y]<map[x][y+1] || y+1<nC && map[x][y+1]==0)// right{// checkbool bHave=false,bNotbigger=false;if (map[x][y]==-1 && map[x][y+1]!=0 || map[x][y+1]==0)// two cases need to check{postmp_stack=position_stack;POSITION psnnext;psnnext.x=x;psnnext.y=y+1;while (postmp_stack.size()){psn=postmp_stack.top();if (psnnext.x==psn.x && psnnext.y==psn.y)// have this point yet {bHave=true;break;}if (map[x][y+1]!=0){if (map[x][y+1]<=map[psn.x][psn.y])// not bigger than before{bNotbigger=true;break;}}postmp_stack.pop();}}// if ok, then save this pointif (!bHave && !bNotbigger){sum+=map[x][y+1];psn.x=x;psn.y=y+1;position_stack.push(psn);if (map[x][y+1]==0){map[x][y+1]=-1;}visit(map,x,y+1);}}if (y-1>=0 && map[x][y]<map[x][y-1] || y-1>=0 && map[x][y-1]==0)// left{// checkbool bHave=false,bNotbigger=false;if (map[x][y]==-1 && map[x][y-1]!=0 || map[x][y-1]==0)// two cases need to check{postmp_stack=position_stack;POSITION psnnext;psnnext.x=x;psnnext.y=y-1;while (postmp_stack.size()){psn=postmp_stack.top();if (psnnext.x==psn.x && psnnext.y==psn.y)// have this point yet {bHave=true;break;}if (map[x][y-1]!=0){if (map[x][y-1]<=map[psn.x][psn.y])// not bigger than before{bNotbigger=true;break;}}postmp_stack.pop();}}// if ok, then save this pointif (!bHave && !bNotbigger){sum+=map[x][y-1];psn.x=x;psn.y=y-1;position_stack.push(psn);if (map[x][y-1]==0){map[x][y-1]=-1;}visit(map,x,y-1);}}// check and save if (sum>=sumres){static bool bFirst=true;if (bFirst){sumres=sum;record_stack=position_stack;bFirst=false;// mark one time}else{// save the shortest and the maximum pathif (position_stack.size()<record_stack.size() || sum>sumres){sumres=sum;record_stack=position_stack;}}}if (position_stack.size()){psn=position_stack.top();if (map[psn.x][psn.y]==-1){sum-=0;map[psn.x][psn.y]=0;// reset} else{sum-=map[psn.x][psn.y];}position_stack.pop();// synif (position_stack.size()==0)// have tried all{cout<<"===Get the answer==="<<endl;return;}}}/*test examples[1]cross.txt2 31 0 42 3 1output:===Get the answer===The biggest weight is: 10The perhaps path is: 1 2 3 0 4[2]cross.txt3 41 0 0 30 2 3 00 1 1 0output:===Get the answer===The biggest weight is: 6The perhaps path is: 1 0 2 3[3]cross.txt2 71 2 13 6 7 12 1114 3 4 5 8 9 10output:===Get the answer===The biggest weight is: 78The perhaps path is: 1 2 3 4 5 6 7 8 9 10 11 12[4]cross.txt5 51 16 15 14 13 2 17 24 23 12 3 18 25 22 11 4 19 20 21 10 5 6 7 8 9 output:===Get the answer===The biggest weight is: 325The perhaps path is: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 [5]cross.txt3 40 0 0 30 2 3 01 1 1 0output:===Get the answer===The biggest weight is: 5The perhaps path is: 0 0 2 3[6]cross.txt3 40 0 0 20 3 4 00 0 1 0output:===Get the answer===The biggest weight is: 10The perhaps path is: 0 0 0 0 1 0 0 2 0 0 3 4[7]cross.txt4 41 0 0 30 2 3 00 1 1 00 0 0 4output:===Get the answer===The biggest weight is: 10The perhaps path is: 1 0 2 3 0 0 4[8]cross.txt6 41 0 0 30 2 3 00 1 1 00 0 0 05 0 0 00 0 0 4output:===Get the answer===The biggest weight is: 15The perhaps path is: 1 0 2 3 0 0 0 0 4 0 0 0 5[9]cross.txt3 41 0 0 2 4 0 0 35 0 0 6output:===Get the answer===The biggest weight is: 21The perhaps path is: 1 0 0 2 3 0 0 4 5 0 0 6*/