UVa 10317 - Equating Equations（DFS）

题意

不改变符号顺序，问能不能交换数字的位置使等式成立。

思路

一开始看数据量这么小果断暴力枚举全排列然后检查，爽快地TLE了。后来想了一下，可以把左边的-号的数字都移到右边，这样两边都是正的数字，只要找出sum / 2的数字组成就行。

这里可以写DFS也可以写DP。

然后就是输出。

取出符号，如果在等号左边，而且符号是+，之后填上我们找到的数字，如果是-，就填上我们找到数字之外的数字（就是移到右边是正的数字）

等号右边相反。

但是需要考虑的情况有点多，主要是越界的问题。代码写得很不优雅，大家参考思路就行。。

代码

  
  
  
  

   
   
   
   
#include <cstdio>
   
   
   
   
#include <stack>
   
   
   
   
#include <set>
   
   
   
   
#include <iostream>
   
   
   
   
#include <string>
   
   
   
   
#include <vector>
   
   
   
   
#include <cctype>
   
   
   
   
#include <algorithm>
   
   
   
   
#include <cstdlib>
   
   
   
   
#include <queue>
   
   
   
   
#include <functional>
   
   
   
   
#include <cstring>
   
   
   
   
#include <string>
   
   
   
   
#include <sstream>
   
   
   
   
#include <map>
   
   
   
   
#include <cmath>
   
   
   
   
#define LL long long
   
   
   
   
#define lowbit(x) ((x) & (-x))
   
   
   
   
#define MP(a, b) make_pair(a, b)
   
   
   
   
#define MS(arr, num) memset(arr, num, sizeof(arr))
   
   
   
   
#define PB push_back
   
   
   
   
#define ROP freopen("input.txt", "r", stdin);
   
   
   
   
const double PI = acos(-1.0);
   
   
   
   
const int INF = 0x3f3f3f3f;
   
   
   
   
using namespace std;
   
   
   
   
const int MAXN = 30 + 5;
   
   
   
   

   
   
   
   
typedef pair<int, int> pii;
   
   
   
   
typedef vector<int>::iterator viti;
   
   
   
   
typedef vector<pii>::iterator vitii;
   
   
   
   

   
   
   
   
int vis[MAXN], succeed, sum, ans[MAXN], cnt;
   
   
   
   
vector<int> num, sym, rnum;
   
   
   
   
char str[10000];
   
   
   
   
bool lpos, rpos;
   
   
   
   

   
   
   
   
void DFS(int curSum, int cur, int curCnt)
   
   
   
   
{
   
   
   
   
 for (int i = cur; i < num.size(); i++)
   
   
   
   
 {
   
   
   
   
 if (succeed) return;
   
   
   
   
 if (curSum + num[i] > sum) return;
   
   
   
   
 if (curSum + num[i] == sum && curCnt + 1 == cnt)
   
   
   
   
 {
   
   
   
   
 ans[curCnt] = i;
   
   
   
   
 succeed = 1;
   
   
   
   
 return;
   
   
   
   
 }
   
   
   
   
 ans[curCnt] = i;
   
   
   
   
 DFS(curSum + num[i], i + 1, curCnt + 1);
   
   
   
   
 }
   
   
   
   
}
   
   
   
   

   
   
   
   
void OutPut()
   
   
   
   
{
   
   
   
   
 if (!succeed) printf("no solution");
   
   
   
   
 else
   
   
   
   
 {
   
   
   
   
 for (int i = 0; i < cnt; i++) vis[ans[i]] = 1;
   
   
   
   
 for (int i = 0; i < num.size(); i++)
   
   
   
   
 if (!vis[i]) rnum.push_back(num[i]);
   
   
   
   
 int i = 0, j = 0, k = 0;
   
   
   
   
 char ch;
   
   
   
   
 if (lpos)
   
   
   
   
 {
   
   
   
   
 ch = sym[j++];
   
   
   
   
 printf("%c", ch);
   
   
   
   
 if (ch == '+') printf("%d", num[ans[i++]]);
   
   
   
   
 if (ch == '-') printf("%d", rnum[k++]);
   
   
   
   
 }
   
   
   
   
 else printf("%d", num[ans[i++]]);
   
   
   
   
 ch = sym[j++];
   
   
   
   
 while (ch != '=')
   
   
   
   
 {
   
   
   
   
 printf(" %c ", ch);
   
   
   
   
 if (ch == '-') printf("%d", rnum[k++]);
   
   
   
   
 if (ch == '+') printf("%d", num[ans[i++]]);
   
   
   
   
 ch = sym[j++];
   
   
   
   
 }
   
   
   
   
 printf(" = ");
   
   
   
   
 if (rpos)
   
   
   
   
 {
   
   
   
   
 ch = sym[j++];
   
   
   
   
 printf("%c", ch);
   
   
   
   
 if (ch == '-') printf("%d", num[ans[i++]]);
   
   
   
   
 if (ch == '+') printf("%d", rnum[k++]);
   
   
   
   
 if (j == sym.size()) return;
   
   
   
   
 }
   
   
   
   
 else printf("%d", rnum[k++]);
   
   
   
   
 if (j == sym.size()) return;
   
   
   
   
 ch = sym[j++];
   
   
   
   
 while (1)
   
   
   
   
 {
   
   
   
   
 printf(" %c ", ch);
   
   
   
   
 if (ch == '-') printf("%d", num[ans[i++]]);
   
   
   
   
 if (ch == '+') printf("%d", rnum[k++]);
   
   
   
   
 if (j == sym.size()) return;
   
   
   
   
 ch = sym[j++];
   
   
   
   
 }
   
   
   
   

   
   
   
   
 }
   
   
   
   
}
   
   
   
   

   
   
   
   
int main()
   
   
   
   
{
   
   
   
   
 //ROP;
   
   
   
   
 int i, j;
   
   
   
   
 char ch;
   
   
   
   
 while (gets(str))
   
   
   
   
 {
   
   
   
   
 MS(vis, 0);
   
   
   
   
 succeed = sum = cnt = 0;
   
   
   
   
 num.clear();
   
   
   
   
 rnum.clear();
   
   
   
   
 sym.clear();
   
   
   
   
 lpos = rpos = false;
   
   
   
   
 int a = 0;
   
   
   
   
 char last = 0;
   
   
   
   

   
   
   
   
 for (i = 0; str[i] != '='; i++)
   
   
   
   
 {
   
   
   
   
 if (isdigit(str[i]))
   
   
   
   
 {
   
   
   
   
 if (last == '+' || last == 0) cnt++;
   
   
   
   
 a = atoi(&str[i]);
   
   
   
   
 num.push_back(a); sum += a;
   
   
   
   
 while (isdigit(str[i])) i++;
   
   
   
   
 }
   
   
   
   
 if (str[i] == '+' || str[i] == '-')
   
   
   
   
 {
   
   
   
   
 if (num.empty()) lpos = true;
   
   
   
   
 last = str[i];
   
   
   
   
 sym.push_back(last);
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 sym.push_back(str[i]);
   
   
   
   
 int temp = num.size(); //用于标记当前的数量
   
   
   
   
 last = 0;
   
   
   
   
 for(; i < strlen(str); i++)
   
   
   
   
 {
   
   
   
   
 if (isdigit(str[i]))
   
   
   
   
 {
   
   
   
   
 if (last == '-') cnt++;
   
   
   
   
 a = atoi(&str[i]);
   
   
   
   
 num.push_back(a); sum += a;
   
   
   
   
 while (isdigit(str[i])) i++;
   
   
   
   
 }
   
   
   
   
 if (str[i] == '+' || str[i] == '-')
   
   
   
   
 {
   
   
   
   
 if (num.size() == temp) rpos = true;
   
   
   
   
 last = str[i];
   
   
   
   
 sym.push_back(last);
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 if (sum & 1)
   
   
   
   
 {
   
   
   
   
 printf("no solution\n");
   
   
   
   
 continue;
   
   
   
   
 }
   
   
   
   
 sum /= 2;
   
   
   
   
 sort(num.begin(), num.end());
   
   
   
   
 DFS(0, 0, 0);
   
   
   
   
 OutPut();
   
   
   
   
 puts("");
   
   
   
   
 }
   
   
   
   
 return 0;
   
   
   
   
}