一、不同路径

 

class Solution {
public:int uniquePaths(int m, int n) {vector<vector<int>> dp(m+1,vector<int>(n+1));dp[0][1] = 1;for(int i = 1;i <= m;i++){for(int j = 1;j <= n;j++){dp[i][j] = dp[i-1][j]+dp[i][j-1];}}return dp[m][n];}
};

二、不同路径II 

 

class Solution {
public:int uniquePathsWithObstacles(vector<vector<int>>& ob) {int m = ob.size();int n = ob[0].size();vector<vector<int>> dp(m+1,vector<int>(n+1));dp[1][0] = 1;for(int i = 1;i <= m;i++){for(int j = 1;j <= n;j++){if(ob[i-1][j-1] == 0) dp[i][j] = dp[i-1][j] + dp[i][j-1];}} return dp[m][n];}
};

 三、礼物的最大价值

class Solution {
public:int maxValue(vector<vector<int>>& grid) {int m = grid.size();int n = grid[0].size();vector<vector<int>> dp(m+1,vector<int>(n+1));for(int i = 1;i <= m;i++){for(int j = 1;j <= n;j++){dp[i][j] = max(dp[i-1][j],dp[i][j-1])+grid[i-1][j-1];}}return dp[m][n];}
};

 四、下降路径最小和

 

 

 

class Solution {
public:int minFallingPathSum(vector<vector<int>>& matrix) {int n = matrix.size();vector<vector<int>> dp(n+1,vector<int> (n+2,INT_MAX));for(int j = 0;j < n+2;j++){dp[0][j] = 0;}for(int i = 1;i <= n;i++){for(int j = 1;j <= n;j++){dp[i][j] = min(dp[i-1][j-1],min(dp[i-1][j],dp[i-1][j+1]))+matrix[i-1][j-1];}}int ret = INT_MAX;for(int i = 1;i <= n;i++)ret = min(ret,dp[n][i]);return ret;}
};

 五、最小路径和

 

此题跟前面差不多，略。

六、地下城游戏 【以[i,j]位置为起点】

        

 

 

class Solution {
public:int calculateMinimumHP(vector<vector<int>>& dungeon) {int m = dungeon.size(),n = dungeon[0].size();vector<vector<int>> dp(m+1,vector(n+1,INT_MAX));dp[m][n-1] = dp[m-1][n] = 1;for(int i = m - 1;i >= 0;i--){for(int j = n - 1;j >=0;j--){dp[i][j] = min(dp[i+1][j],dp[i][j+1]) - dungeon[i][j];dp[i][j] = max(1,dp[i][j]);}}return dp[0][0];}
};