为什么 dist[x]<= d (v,x)

这个问题涉及到图论中的最短路径问题。假设我们要求从顶点v到顶点x的最短路径，其中dist[x]表示顶点x到起点v的最短距离，d(v,x)表示v到x的距离。那么，dist[x]<= d(v,x)的原因是因为在计算最短路径时，我们会对已经访问过的顶点进行标记，以确保每个顶点的最短路径只会被计算一次。因此，当我们到达顶点x时，它的最短路径已经被更新为它到起点v的最短距离，因此dist[x]<= d(v,x)。

相关问题

#include <stdio.h> #include <stdlib.h> #define MAXN 1000 #define INF 0x3f3f3f3f int n, m; int graph[MAXN][MAXN]; int path[MAXN]; int dist; void dfs(int u, int depth, int distance, int visited[]) { if (depth == n) { if (graph[u][1] != INF && distance + graph[u][1] < dist) { dist = distance + graph[u][1]; for (int i = 1; i <= n; i++) { printf("%d ", path[i]); } printf("1\n"); } return; } for (int v = 2; v <= n; v++) { if (!visited[v] && graph[u][v] != INF) { visited[v] = 1; path[depth+1] = v; dfs(v, depth+1, distance+graph[u][v], visited); visited[v] = 0; } } } int main() { scanf("%d %d", &n, &m); for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { graph[i][j] = INF; } } for (int i = 1; i <= m; i++) { int x, y, z; scanf("%d,%d,%d", &x, &y, &z); graph[x][y] = graph[y][x] = z; } for (int k = 1; k <= n; k++) { for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (graph[i][k] != INF && graph[k][j] != INF && graph[i][k] + graph[k][j] < graph[i][j]) { graph[i][j] = graph[i][k] + graph[k][j]; } } } } int visited[MAXN] = {0}; visited[1] = 1; path[1] = 1; dist = INF; dfs(1, 1, 0, visited); printf("%d\n", dist); return 0; }修改改代码，使其只显示最短路径

可以在dfs函数中加入一个变量来记录当前的最短路径长度，每次更新路径时判断是否比当前最短路径更短。同时，在找到最短路径后就直接return，不再继续搜索其他路径。

修改后的代码如下：

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

#define MAXN 1000
#define INF 0x3f3f3f3f

int n, m;
int graph[MAXN][MAXN];
int path[MAXN];
int dist;
int shortest_dist = INF;

void dfs(int u, int depth, int distance, int visited[]) {
    if (depth == n) {
        if (graph[u][1] != INF && distance + graph[u][1] < shortest_dist) {
            shortest_dist = distance + graph[u][1];
            dist = distance + graph[u][1];
            for (int i = 1; i <= n; i++) {
                path[i] = i;
            }
            path[n+1] = 1;
        }
        return;
    }
    for (int v = 2; v <= n; v++) {
        if (!visited[v] && graph[u][v] != INF) {
            visited[v] = 1;
            path[depth+1] = v;
            dfs(v, depth+1, distance+graph[u][v], visited);
            visited[v] = 0;
        }
    }
}

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            graph[i][j] = INF;
        }
    }
    for (int i = 1; i <= m; i++) {
        int x, y, z;
        scanf("%d,%d,%d", &x, &y, &z);
        graph[x][y] = graph[y][x] = z;
    }
    for (int k = 1; k <= n; k++) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                if (graph[i][k] != INF && graph[k][j] != INF && graph[i][k] + graph[k][j] < graph[i][j]) {
                    graph[i][j] = graph[i][k] + graph[k][j];
                }
            }
        }
    }
    int visited[MAXN] = {0};
    visited[1] = 1;
    path[1] = 1;
    dfs(1, 1, 0, visited);
    printf("%d\n", dist);
    for (int i = 1; i <= n+1; i++) {
        printf("%d ", path[i]);
    }
    printf("\n");
    return 0;
}

为什么d(v, x) + d(x, u) = d(v,u) < dist[u]

这是因为根据三角不等式，d(v, x) + d(x, u) >= d(v, u)，同时dist[u]是从起点s到u的最短距离，因此d(v, u) <= dist[u]。因此，如果d(v, x) + d(x, u) < dist[u]，那么d(v, u) < dist[u]，与dist[u]为最短距离的定义不符。因此，d(v, x) + d(x, u) >= d(v, u)必须成立。