Solution to a problem P2327 [SCOI2005] Minesweeper

Solution to a problem P2327 [SCOI2005] Minesweeper

Topic Link

Universal set of water problemsVery suitable for beginners to practice search

Look problem solutions are used in a variety of deities simple way, I LaijiangjiangviolenceSearch practice ....

For me I think that sprout new approach easier for you to understand search

To define the state of the first search, we define dfs (x, p) when x bit is p, there will be a state

Then we discuss the first case x + 1 bitsOnly two casesThen there will be a transfer

Secondly, when x is n, if true, then ++ ans, so with the boundary conditions

Finally, copy and paste the code you got 100 points and found that it's a bit like dynamic programming

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define ll long long
#define N 100005
int a[N], b[N], n;//a是左边那排，b是每一个格子是否放雷
ll ans;
void dfs(int x, int p) //第x个格子放 p个雷
{
    b[x] = p;
    if (x == n) //边界
    {
        if (b[x] + b[x - 1] == a[x]) //第n个格子放 p个雷 成立
            ans++;
        return;
    }
    for (int i = 0; i <= 1; i++) //讨论第x + 1个格子放不放雷
        if (b[x] + b[x - 1] + i == a[x] && b[x] + i <= a[x + 1])
            dfs(x + 1, i); //转移（滑稽
    b[x] = 0;
}
int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    dfs(1, 1);
    dfs(1, 0);
    cout << ans;
    return 0;
}

19.08.18