nyoj 927 The partial sum problem（dfs）

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One day,Tom’s girlfriend give him an array A which contains N integers and asked him:Can you choose some integers from the N integers and the sum of them is equal to K. 
 

 

输入

There are multiple test cases.
Each test case contains three lines.The first line is an integer N(1≤N≤20),represents the array contains N integers. The second line contains N integers,the ith integer represents A[i](-10^8≤A[i]≤10^8).The third line contains an integer K(-10^8≤K≤10^8).

 

输出

If Tom can choose some integers from the array and their them is K,printf ”Of course,I can!”; other printf ”Sorry,I can’t!”.

 

样例输入

4
1 2 4 7
13
4
1 2 4 7
15

 

样例输出

Of course,I can!
Sorry,I can‘t!

 

两种方法：

第一种直接回溯dfs

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<math.h>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 #include <stack>
15 using namespace std;
16 #define PI acos(-1.0)
17 #define max(a,b) (a) > (b) ? (a) : (b)
18 #define min(a,b) (a) < (b) ? (a) : (b)
19 #define ll long long
20 #define eps 1e-10
21 #define MOD 1000000007
22 #define N 26
23 #define inf 1e12
24 int n,m,flag;
25 int a[N];
26 int vis[N];
27 void dfs(int now,int num){
28    if(num>=m){
29          if(num==m){
30             flag=1;
31          }
32          return;
33    }
34    for(int i=now;i<n;i++){
35       if(!vis[i]){
36          vis[i]=1;
37          dfs(i+1,num+a[i]);
38          if(flag){
39             return;
40          }
41          vis[i]=0;
42       }
43    }
44 }
45 int main()
46 {
47    while(scanf("%d",&n)==1){
48       for(int i=0;i<n;i++){
49          scanf("%d",&a[i]);
50       }
51       scanf("%d",&m);
52       memset(vis,0,sizeof(vis));
53       flag=0;
54       dfs(0,0);
55       if(flag){
56          printf("Of course,I can!\n");
57       }else{
58          printf("Sorry,I can‘t!\n");
59       }
60    }
61     return 0;
62 }

View Code

 

第二种类似01背包思想的dfs

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<math.h>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 #include <stack>
15 using namespace std;
16 #define PI acos(-1.0)
17 #define max(a,b) (a) > (b) ? (a) : (b)
18 #define min(a,b) (a) < (b) ? (a) : (b)
19 #define ll long long
20 #define eps 1e-10
21 #define MOD 1000000007
22 #define N 26
23 #define inf 1e12
24 int n,m;
25 int a[N];
26 bool dfs(int cur,int num){
27    if(num>=m){
28       if(num==m){
29          return true;
30       }
31       return false;
32    }
33    if(cur>=n) return false;
34    if(dfs(cur+1,num+a[cur])) return true;
35    return dfs(cur+1,num);
36 
37 }
38 int main()
39 {
40    while(scanf("%d",&n)==1){
41       for(int i=0;i<n;i++){
42          scanf("%d",&a[i]);
43       }
44       scanf("%d",&m);
45       if(dfs(0,0)){
46          printf("Of course,I can!\n");
47       }else{
48          printf("Sorry,I can‘t!\n");
49       }
50    }
51     return 0;
52 }

View Code

 

 

nyoj 927 The partial sum problem（dfs）

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原文地址：http://www.cnblogs.com/UniqueColor/p/4984753.html