算法笔记上机训练实战指南第8章提高篇(2)--搜索专题学习笔记

8.1 深度优先搜索(DFS)

PAT A1103 Integer Factorization (30分)

The

Input Specification:

Each input file contains one test case which gives in a line the three positive integers

Output Specification:

For each case, if the solution exists, output in the format:

N = n[1]^P + ... n[K]^P

where n[i] (i = 1, ..., K) is the i-th factor. All the factors must be printed in non-increasing order.

Note: the solution may not be unique. For example, the 5-2 factorization of 169 has 9 solutions, such as $1, or 1, or more. You must output the one with the maximum sum of the factors. If there is a tie, the largest factor sequence must be chosen -- sequence {, } is said to be larger than {, } if there exists 1 such that$

If there is no solution, simple output Impossible.

Sample Input 1:

169 5 2

Sample Output 1:

169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2

Sample Input 2:

169 167 3

Sample Output 2:

Impossible

#include<cstdio>
#include<vector>
using namespace std;
int n,k,p,maxfacSum = - 1;
vector<int> temp,ans,fac;
int power(int x){
    int ans = 1;
    for(int i = 0; i < p;i++){
        ans *= x;
    }
    return ans;
}
void init(){
    int temp = 0,i = 0;
    while(temp <= n){
        fac.push_back(temp);
        temp = power(++i);
    }
}
void DFS(int index,int nowK,int sum,int facSum){
    if(nowK == k && sum == n){
        if(facSum > maxfacSum){
            ans = temp;
            maxfacSum = facSum;
        }
        return;
    }
    if(nowK > k || sum > n)
        return;
    if(index-1 >= 0){
        temp.push_back(index);
        DFS(index,nowK+1,sum+fac[index],facSum+index);
        temp.pop_back();
        DFS(index-1,nowK,sum,facSum);
    }
}
int main(){
    scanf("%d%d%d",&n,&k,&p);
    init();
    DFS(fac.size()-1,0,0,0);
    if(maxfacSum == -1){
        printf("Impossible
");
    }else{
        printf("%d = %d^%d",n,ans[0],p);
        for(int i =1;i<ans.size();i++){
            printf(" + %d^%d",ans[i],p);
        }
    }
    return 0;
}

8.2 广度优先搜索(BFS)

PAT A1091 Acute Stroke (30分)

One important factor to identify acute stroke (急性脑卒中) is the volume of the stroke core. Given the results of image analysis in which the core regions are identified in each MRI slice, your job is to calculate the volume of the stroke core.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 positive integers:

Then

Figure 1

Output Specification:

For each case, output in a line the total volume of the stroke core.

Sample Input:

Sample Output:

#include<cstdio>
#include<queue>
using namespace std;
struct node{
    int x,y,z;
}Node;
int m,n,l,T;
int pixel[1290][130][61];
bool inq[1290][130][61] = {false};
int X[6]={0,0,0,0,1,-1};
int Y[6]={0,0,1,-1,0,0};
int Z[6]={1,-1,0,0,0,0};
bool judge(int x,int y,int z){
    if(x<0 || x>=m || y<0 || y>=n||z<0 || z>=l){
        return false;
    }
    if(inq[x][y][z] == true || pixel[x][y][z] == 0)
        return false;
    return true;
}
int BFS(int x,int y,int z){
    int tot = 0;
    queue<node> Q;
    Node.x = x,Node.y = y,Node.z = z;
    Q.push(Node);
    inq[x][y][z] = true;
    while(!Q.empty()){
        node top = Q.front();
        Q.pop();
        tot++;
        for(int i=0;i<6;i++){
            int newX = top.x + X[i];
            int newY = top.y + Y[i];
            int newZ = top.z + Z[i];
            if(judge(newX,newY,newZ)){
                Node.x = newX,Node.y = newY,Node.z = newZ;
                Q.push(Node);
                inq[newX][newY][newZ] = true;
            }
        }
    }
    if(tot >= T)
        return tot;
    else
        return 0;
}
int main(){
    scanf("%d%d%d%d",&m,&n,&l,&T);
    for(int z=0;z<l;z++){
        for(int x=0;x<m;x++){
            for(int y = 0;y <n;y++){
                scanf("%d",&pixel[x][y][z]);
            }
        }
    }
    int ans = 0;
    for(int z=0;z<l;z++){
        for(int x=0;x<m;x++){
            for(int y = 0;y <n;y++){
                if(pixel[x][y][z] == 1 && inq[x][y][z] == false){
                    ans += BFS(x,y,z);
                }
            }
        }
    }
    printf("%d
",ans);
    return 0;
}