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36 lines
598 B
36 lines
598 B
#include <bits/stdc++.h>
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using namespace std;
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const int INF = 0x3f3f3f3f;
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const int N = 110;
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int a[N][N];
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int n, m;
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int f[N][N];
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// dfs解法
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/*
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2 3
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1 3 5
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2 4 6
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答案：
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13
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*/
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int dfs(int x, int y) {
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if (f[x][y]) return f[x][y];
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if (x > n || y > m) return INF;
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if (x == n && y == m) return a[n][m];
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return f[x][y] = a[x][y] + min(dfs(x + 1, y), dfs(x, y + 1));
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}
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int main() {
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cin >> n >> m;
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++)
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cin >> a[i][j];
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printf("%d\n", dfs(1, 1));
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return 0;
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} |