用c语言由邻接矩阵输出任意两个节点长度小于等于n的路径的数目

_loss = 0.0
total_acc = 0.0
with torch.no_grad():
for inputs, labels in dataloader:
inputs = {key: inputs[key].to(device) for key in inputs}
labels = labels.to(device)
下面是用C语言实现的邻接矩阵输出任意两个节点长度小于等于n的路径的数目的代码：

#include<stdio.h>
#include<stdlib.h>

#define MAXN 100

int graph[MAXN][MAXN]; outputs = model(inputs)
            loss = criterion(outputs, labels)
            total_loss += loss.item() * len(labels)
            preds = torch  // 邻接矩阵
int n;                  // 节点数

int count_paths(int u, int v, int len) {
.argmax(outputs, dim=1)
            total_acc += torch.sum(preds == labels).item()
    avg_loss = total_loss / len(d    if (len == 0) {
        if (u == v) {
            return 1;   // u到v的长度ataloader.dataset)
    avg_acc = total_acc / len(dataloader.dataset)
    return avg_loss, avg_acc

# 定义评估函数为0的路径数为1
        } else {
            return 0;
        }
    } else {
        int count = 0
def evaluate_test(model, dataloader, device):
    model.eval()
    total_acc = 0.0
    with torch.no;
        for (int i = 0; i < n; i++) {
            if (graph[u][i]) {
                count +=_grad():
        for inputs, labels in dataloader:
            inputs = {key: inputs[key].to(device) for key in inputs count_paths(i, v, len - 1); // 递归计算长度为len-1的路径数
            }
       }
            labels = labels.to(device)
            outputs = model(inputs)
            preds = torch.argmax(outputs, dim=1)
            total_acc }
        return count;
    }
}

int main() {
    // 读入邻接矩阵
    scanf("%d", &n += torch.sum(preds == labels).item()
    avg_acc = total_acc / len(dataloader.dataset)
    return avg_acc

);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n最后，我们可以开始训练模型并进行评估：

# 初始化模型和优化器
device =; j++) {
            scanf("%d", &graph[i][j]);
        }
    }

    // 输出任意两个节点长度小 torch.device('cuda' if torch.cuda.is_available() else 'cpu')
model = TextClassifier(len(class_list)).to(device)
criterion于等于n的路径的数目
    int ans = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            for (int len = 1; = nn.CrossEntropyLoss()
optimizer = optim.Adam(model.parameters(), lr=1e-5)

# 训练模型
num_epochs = len <= n; len++) {
                ans += count_paths(i, j, len);
            }
        }
    }
    printf("%d\n 5
best_valid_acc = 0.0
for epoch in range(num_epochs):
    train_loss, train_acc = train(model,", ans);

    return 0;
}

这个算法的时间复杂度为$O(n^3)$，可以 train_loader, optimizer, criterion, device)
valid_loss, valid_acc = evaluate(model, valid_loader, criterion, device)
print处理节点数比较小的图。如果节点数比较大，可以考虑使用其他更高效的算法，如Floyd算法或Dijkstra算法。