memset(v, 0x3f

`memset` 是 C/C++ 中的一个库函数，用于将一段内存区域初始化为特定的值。在您的示例中，`memset(v, 0x3f, size)` 将把变量 `v` 所指向的大小为 `size` 的内存区域设置为十六进制数 `0x3f`（即二进制的 `111111`）。

这个操作通常用于预填充数组或者标记某些位，因为 `0x3f` 被转换成二进制后，每一位都是 1，可以快速地查看哪些位置已经被设置过。

**示例代码:**

#include <string.h> // 包含 memset 函数

int main() {
    int v[10]; // 假设这是一个整型数组
    size_t size = sizeof(v); // 获取数组大小

    memset(v, 0x3f, size); // 初始化 v 中的所有元素为 0x3f

    // 现在 v 的每个元素都被设置为 0x3f (二进制表示)
    // 你可以通过遍历数组来查看这些值

    return 0;
}

相关问题

memset 0x3f

引用中提到，memset(dis, 0x3f, sizeof(dis))的意思是将数组dis中的每个元素都设置为0x3f（十进制为63）。 这里的0x3f表示一个特定的值，可以根据具体情况进行设置。在引用中提到，0x3f3f3f3f是一个常用的十六进制值，在一些算法中常用于表示无穷大或者某种特殊情况。在引用中提到，memset是一个标准库函数，用于将一块内存的每个字节设置为特定的值。所以，memset 0x3f的意思就是使用memset函数将特定的值0x3f设置到内存中的每个字节中。<span class="em">1</span><span class="em">2</span><span class="em">3</span>
#### 引用[.reference_title]
- *1* *3* [memset( dis,0x3f ,sizeof(dis) )](https://blog.csdn.net/m0_73813813/article/details/131494973)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 50%"]
- *2* [图论中的0x3f和memset使用注意事项（较详细）](https://blog.csdn.net/weixin_45717583/article/details/117335935)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 50%"]
[ .reference_list ]

对以下c++代码进行改编但不改变代码准确性：#include <iostream> #include <cstdio> #include <cmath> #include <cstring> using namespace std; typedef long long LL; const int N = 105, M = 2505, L = 20; const LL INF = 1e18; int n, m, K, l; LL d[N][N], w[N][N], g[L][N][N], f[N], t[N]; struct E{ int u, v, w; } e[M]; int main() { memset(g, 0x3f, sizeof g); scanf("%d%d%d", &n, &m, &K); l = log2(K); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) if (i != j) d[i][j] = INF; for (int i = 1; i <= m; i++) { scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w); d[e[i].u][e[i].v] = min(d[e[i].u][e[i].v], (LL)e[i].w); } for (int k = 1; k <= n; k++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) d[i][j] = min(d[i][j], d[i][k] + d[k][j]); for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { w[i][j] = d[i][j]; for (int k = 1; k <= m; k++) w[i][j] = min(w[i][j], d[i][e[k].u] - e[k].w + d[e[k].v][j]); g[0][i][j] = w[i][j]; } } for (int c = 1; c <= l; c++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) for (int k = 1; k <= n; k++) g[c][i][j] = min(g[c][i][j], g[c - 1][i][k] + g[c - 1][k][j]); for (int i = 1; i <= n; i++) f[i] = d[1][i]; for (int c = 0; c <= l; c++) { if (K >> c & 1) { for (int i = 1; i <= n; i++) t[i] = f[i]; memset(f, 0x3f, sizeof f); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) f[i] = min(f[i], t[j] + g[c][j][i]); } } printf("%lld\n", f[n]); return 0; }

#include <iostream>
#include <cstdio>
#include <cmath>
#include <>
using namespace std; typedef long long LL const int N = 105, M = 2505, L = 20;
const LL INF = 1e18; int n, m, K, l;
LL d[N][N], w[N][N], g[L][N][N], f[N], t[N]; struct E{
int u, v, w;
} e[M]; int main() {
memset(g, 0x3f, sizeof g);
scanf("%d%d%d", &n, &m, &K);
l = log2(K);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) if (i != j) d[i][j] = INF;
for (int i = 1; i <= m; i++) {
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
d[e[i].u][e[i].v] = min(d[e[i].u][e[i].v], (LL)e[i].w);
}
for (int k = 1; k <= n; k++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
w[i][j] = d[i][j];
for (int k = 1; k <= m; k++)
w[i][j] = min(w[i][j], d[i][e[k].u] - e[k].w + d[e[k].v][j]);
g[0][i][j] = w[i][j];
}
}
for (int c = 1; c <= l; c++)
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
for (int k = 1; k <= n; k++)
g[c][i][j] = min(g[c][i][j], g[c - 1][i][k] + g[c - 1][k][j]);
for (int i = 1; i <= n; i++) f[i] = d[1][i];
for (int c = 0; c <= l; c++) {
if (K >> c & 1) {
for (int i = 1; i <= n; i++) t[i] = f[i];
memset(f, 0x3f, sizeof f);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++) f[i] = min(f[i], t[j] + g[c][j][i]);
}
}
printf("%lld\n", f[n]);
return 0;
}