add lanqiao216

test99-4-5.cpp

@@ -1,3 +1,4 @@
+// 蓝桥3820 混境之地
 #include<bits/stdc++.h>
 using namespace std;
 using ll = long long;
@@ -35,7 +36,7 @@ bool dfs(int x, int y, int t){
     return dp[x][y][t] = false;
 }
 int main(){
-    memset(dp, -1, sizeof dp);
+    memset(dp, -1, sizeof dp);  // 记忆化搜索
     cin >> n >> m >> k;
     cin >> sx >> sy >> fx >> fy;
     for(int i = 1; i <= n; i++){

test99-4-6.cpp

@@ -0,0 +1,49 @@
+// lanqiao216 地宫寻宝
+#include<bits/stdc++.h>
+using namespace std;
+using ll = long long;
+const ll p = 1e9 + 7;
+const int N = 55;
+int n, m, k, c[N][N];
+int dx[] = {0, 1};
+int dy[] = {1, 0};
+int dp[N][N][15][15];
+
+bool inmp(int x, int y){
+    return x >= 1 && x <= n && y >= 1 && y <= m;
+}
+
+// 表示从(1,1)到(x, y), 总共有cnt件宝贝, 且最大值为max的方案数
+ll dfs(int x, int y, int mx, int cnt){
+    if(x == n && y == m) return (ll)(cnt == k);
+    if(dp[x][y][mx][cnt] != -1) return dp[x][y][mx][cnt];
+
+    ll res = 0;
+    for(int i = 0; i < 2; i++){
+        int nx = x + dx[i], ny = y + dy[i];
+        if(!inmp(nx, ny)) continue;
+        // 拿宝贝
+        if(c[nx][ny] > mx && cnt < k) res = (res + dfs(nx, ny, c[nx][ny], cnt + 1)) % p;
+        //不拿宝贝
+        res = (res + dfs(nx, ny, mx, cnt)) % p;
+    }
+    return dp[x][y][mx][cnt] = res;
+}
+
+int main(){
+    memset(dp, -1, sizeof dp);
+    cin >> n >> m >> k;
+    for(int i = 1; i <= n; i++){
+        for(int j = 1; j <= m; j++){
+            cin >> c[i][j];
+            c[i][j]++;
+        }
+    }
+    cout << (dfs(1,1,0,0) + dfs(1,1,c[1][1],1)) % p << '\n';
+    return 0;
+}
+/* test samples 1 -> 2
+2 2 2
+1 2
+2 1
+*/