LeetCode刷题笔记 LCP 07 传递信息

菜鸡双栈解法（单纯记录）

class Solution {
public:
    int numWays(int n, vector<vector<int>>& relation, int k) {
        stack<vector<int>> pre,cur;
        sort(relation.begin(),relation.end());
        int i=0,res=0;
        while(i<relation.size()){
            if(relation[i][0]==0) pre.push(relation[i++]);
            else break;
        }
        for(int i=2;i<=k;i++){
            while(!pre.empty()){
                int right=pre.top()[1];
                pre.pop();
                int j=0;
                while(j<relation.size()&&relation[j][0]<right) j++;
                while(j<relation.size()&&relation[j][0]==right){
                    cur.push(relation[j]);
                    j++;
                }
            }
            swap(pre,cur);
        }        
        while(!pre.empty()){
            int right=pre.top()[1];
            pre.pop();
            if(right==n-1) res++;
        }
        return res;       
    }
};

大神解法（重要）

$dp[k][n]$表示第几轮到达编号 $n$ 有几条途径。

第 $i$ 轮能到达编号 $n$ 的点的路径数为第 $i-1$ 轮到达对应路径起点的路径数之和
我们每一轮 $i$ 遍历所有的关系，那么比如对应关系 $[0, 2]$
我们应该累加上一轮能够到达起点，即编号0位置的路径数，即 $dp[i - 1][v[0]]$

class Solution {
public:
    int dp[6][10];
    int numWays(int n, vector<vector<int>>& relation, int k) {
        dp[0][0] = 1;  / 默认一开始即第0轮到达起点
        for(int i = 1; i <= k; i += 1)
            for(auto v : relation)
                dp[i][v[1]] += dp[i - 1][v[0]]; 
        return dp[k][n - 1];
    }
};

流程打印

class Solution {
public:
    int dp[6][10];
    int numWays(int n, vector<vector<int>>& relation, int k) {
        dp[0][0] = 1; 
        for(int i = 1; i <= k; i += 1){
            cout<<"i = "<<i<<endl;
            for(auto v : relation){
                int a=dp[i][v[1]];
                cout<<"dp["<<i<<"]["<<v[1]<<"] + dp["<<i-1<<"]["<<v[0]<<"] = ";
                dp[i][v[1]] += dp[i - 1][v[0]];
                cout<<"dp["<<i<<"]["<<v[1]<<"]"<<endl;
                cout<<a<<" + "<<dp[i - 1][v[0]]<<" = "<<dp[i][v[1]]<<endl;                                
            } 
            cout<<endl;
        }
        return dp[k][n - 1];            
    }
};

输入
5
[[0,2],[2,1],[3,4],[2,3],[1,4],[2,0],[0,4]]
3
输出
3
预期结果
3

i = 1
dp[1][2] + dp[0][0] = dp[1][2]
0 + 1 = 1
dp[1][1] + dp[0][2] = dp[1][1]
0 + 0 = 0
dp[1][4] + dp[0][3] = dp[1][4]
0 + 0 = 0
dp[1][3] + dp[0][2] = dp[1][3]
0 + 0 = 0
dp[1][4] + dp[0][1] = dp[1][4]
0 + 0 = 0
dp[1][0] + dp[0][2] = dp[1][0]
0 + 0 = 0
dp[1][4] + dp[0][0] = dp[1][4]
0 + 1 = 1

i = 2
dp[2][2] + dp[1][0] = dp[2][2]
0 + 0 = 0
dp[2][1] + dp[1][2] = dp[2][1]
0 + 1 = 1
dp[2][4] + dp[1][3] = dp[2][4]
0 + 0 = 0
dp[2][3] + dp[1][2] = dp[2][3]
0 + 1 = 1
dp[2][4] + dp[1][1] = dp[2][4]
0 + 0 = 0
dp[2][0] + dp[1][2] = dp[2][0]
0 + 1 = 1
dp[2][4] + dp[1][0] = dp[2][4]
0 + 0 = 0

i = 3
dp[3][2] + dp[2][0] = dp[3][2]
0 + 1 = 1
dp[3][1] + dp[2][2] = dp[3][1]
0 + 0 = 0
dp[3][4] + dp[2][3] = dp[3][4]
0 + 1 = 1
dp[3][3] + dp[2][2] = dp[3][3]
0 + 0 = 0
dp[3][4] + dp[2][1] = dp[3][4]
1 + 1 = 2
dp[3][0] + dp[2][2] = dp[3][0]
0 + 0 = 0
dp[3][4] + dp[2][0] = dp[3][4]
2 + 1 = 3