[Machine test real problem] [2015]65. Monk Fetching Water

Topics

Solution to the method of thinking back to the Moon

回朔法即每进行一步，都试图在当前部分解的基础上扩大该部分解。扩大时，首先检查扩大后是否违反了约束条件，若不违反，则扩大之，然后继续在此基础上按照类似的方法进行，直至成为完整解；若违反，则放弃该步以及它所能生成的部分解，然后按照类似的方法尝试其他可能的扩大方式，直到尝试了所有的扩大方式。请输入和尚1的空闲时间：0 1 0 1 0 0 0请输入和尚2的空闲时间：1 0 0 0 0 1 0请输入和尚3的空闲时间：0 0 1 0 0 0 1请输入和尚4的空闲时间：0 0 0 0 1 0 0请输入和尚5的空闲时间：1 0 0 1 0 1 0请输入和尚6的空闲时间：0 1 0 0 1 0 0请输入和尚7的空闲时间：0 0 1 0 0 1 1

Code

/* ---------------------------------------* Date: 2015-07-06* sjf0115* title: Monk Carrying Water * Source: Machine Test real problem------------------------- ----------------*/#include <iostream>#include <string>#include <vector>#include <stack>#include <algorithm>using namespace STD;voidDrawingwater ( vector<vector<int> >&spare,intIndexint&count, vector<int>&week, vector<bool>&visited) {intIif(Index = =7) {++count;cout<<"Solution"<<count<<endl; for(i =0; I <7; ++i) {cout<<"Week"<<i+1<<"Monk"<<week[i]+1<<"Water"<<endl; }//for cout<<endl;return; }//if  for(i =0; I <7; ++i) {Week[index] = i;//Judge if Monk J has already picked water and Monk week N is free if(!visited[i] && spare[i][index] = =1) {Visited[i] =true; Drawingwater (spare,index+1, count,week,visited); Visited[i] =false; }//if}//for}intMain () { vector<vector<int> >Spare7, vector<int>(7,0));//freopen ("C:\\users\\administrator\\desktop\\acm.txt", "R", stdin);  for(inti =0; I <7; ++i) {//Initialize Monk's free time  for(intj =0; J <7; ++j) {Cin>>spare[i][j]; }//for}//for intCount =0; vector<int>Week7,0); vector<bool>Visited (7,false); Drawingwater (Spare,0, count,week,visited);cout<<"Total"<<count<<"Kind of Solution"<<endl;return 0;}

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[Machine test real problem] [2015]65. Monk Fetching Water