DP
//dp[i][j] = min(dp[i + 1][j], dp[i + 1][j + 1]) + a[i][j]; #include <iostream> #include <algorithm> using namespace std; int a[101][101]; int dp[101][101]; int n; int main() { int num; cin >> num; while (num--) { cin >> n; fill(a[0], a[0] + 101 * 101, 0); fill(dp[0], dp[0] + 101 * 101, 0); for (int i = 0; i < n; ++i) { for (int j = 0; j <= i; ++j) { cin >> a[i][j]; } } for (int i = n - 1; i >= 0; --i) { for (int j = 0; j <= i; ++j) { dp[i][j] = min(dp[i + 1][j], dp[i + 1][j + 1]) + a[i][j]; } } cout << dp[0][0] << endl; } return 0; }
DFS
#include <iostream> #include <algorithm> using namespace std; int a[101][101]; int n; int _min; void DFS(int index, int sum, int j) { if (index > n) { return; } if (index == n) { if (sum < _min) { _min = sum; } return; } DFS(index + 1, sum + a[index][j], j); if (index != 0) { DFS(index + 1, sum + a[index][j + 1], j + 1); } } int main() { int num; cin >> num; while (num--) { cin >> n; _min = 100000000; fill(a[0], a[0] + 101 * 101, 0); for (int i = 0; i < n; ++i) { for (int j = 0; j <= i; ++j) { cin >> a[i][j]; } } DFS(0, 0, 0); cout << _min << endl; } return 0; }
做半天a不了,后来发现数塔里的向左对应二维数组的向下,向右对应二维数组的向右下。。。
using namespace std; int tower[105][105]; int dp[105][105]; int minn, lev; int main() { int T; scanf(“%d”, &T); while (T > 0) { scanf(“%d”, &lev); fill(dp[0], dp[0] + 105 * 105, 0); for (int i = 1;i <= lev;i++) { for (int j = 1;j <= i;j++) { scanf(“%d”, &tower[i][j]); } } // dp入口 for (int i = 1;i <= lev;i++) { dp[lev][i] = tower[lev][i]; } for (int i = lev - 1;i >= 1;i–) { for (int j = i;j >= 1;j–) { dp[i][j] = min(dp[i + 1][j], dp[i + 1][j + 1]) + tower[i][j]; } } printf(“%d\n”, dp[1][1]); T–; } return 0; } 和dfs有什么关系?
感觉这题dfs比较好写
def dfs(L, i, j, N, sum): global min if i == N-1: a = sum + L[i][j] min = a if a < min else min return dfs(L, i+1, j, N, sum+L[i][j]) dfs(L, i+1, j+1, N, sum+L[i][j]) C = int(input()) for n in range(C): N = int(input()) L = [] for i in range(N): L.append([int(i) for i in input().split()]) min = 0xffffffff dfs(L, 0, 0, N, 0) print(min)
这题有毒。。。 样例数据复制进Clion之后,塔的第四行开始读到的数据全是0,交到OJ上就AC了
自底向上dp
using namespace std;
int main() { int i,j,N,n,a[105][105];scanf(“%d”,&N); while(N–) { scanf(“%d”,&n); for(i=0;i<n;++i) for(j=0;j<=i;++j) scanf(“%d”,&a[i][j]); for(i=n-2;i>=0;–i) for(j=0;j<=i;++j) a[i][j]+=min(a[i+1][j],a[i+1][j+1]); printf(“%d\n”,a[0][0]); }
return 0;
}
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DP
DFS
做半天a不了,后来发现数塔里的向左对应二维数组的向下,向右对应二维数组的向右下。。。
include
include
include
using namespace std;
int tower[105][105];
int dp[105][105];
int minn, lev;
int main() {
int T;
scanf(“%d”, &T);
while (T > 0) {
scanf(“%d”, &lev);
fill(dp[0], dp[0] + 105 * 105, 0);
for (int i = 1;i <= lev;i++) {
for (int j = 1;j <= i;j++) {
scanf(“%d”, &tower[i][j]);
}
}
// dp入口
for (int i = 1;i <= lev;i++) {
dp[lev][i] = tower[lev][i];
}
for (int i = lev - 1;i >= 1;i–) {
for (int j = i;j >= 1;j–) {
dp[i][j] = min(dp[i + 1][j], dp[i + 1][j + 1]) + tower[i][j];
}
}
printf(“%d\n”, dp[1][1]);
T–;
}
return 0;
}
和dfs有什么关系?
感觉这题dfs比较好写
这题有毒。。。
样例数据复制进Clion之后,塔的第四行开始读到的数据全是0,交到OJ上就AC了
自底向上dp
include
using namespace std;
int main()
{
int i,j,N,n,a[105][105];scanf(“%d”,&N);
while(N–)
{
scanf(“%d”,&n);
for(i=0;i<n;++i)
for(j=0;j<=i;++j)
scanf(“%d”,&a[i][j]);
for(i=n-2;i>=0;–i)
for(j=0;j<=i;++j)
a[i][j]+=min(a[i+1][j],a[i+1][j+1]);
printf(“%d\n”,a[0][0]);
}
}
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