POJ 1502 MPI Maelstrom：Dijkstra 單源最短路

題意

給定圖的邊權（下三角矩陣輸入），從節點 1 出發，求到最遠節點的最短路徑長度（即所有節點最短路的最大值）。

解法

標準 Dijkstra 算法。注意輸入格式：第 i 行有 i-1 個數，對應邊 (i, j) (j < i)；輸入可能包含非數字字符需跳過。N ≤ 100，鄰接矩陣儲存即可。

代碼 (C++)

#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int MAXN = 110;
int n, dist[MAXN], graph[MAXN][MAXN];
bool vis[MAXN];

void dijkstra(int s) {
    fill(dist, dist + MAXN, INF);
    fill(vis, vis + MAXN, false);
    dist[s] = 0;
    for (int iter = 1; iter <= n; iter++) {
        int u = -1;
        for (int i = 1; i <= n; i++)
            if (!vis[i] && (u == -1 || dist[i] < dist[u])) u = i;
        if (u == -1) break;
        vis[u] = true;
        for (int v = 1; v <= n; v++)
            if (dist[v] > dist[u] + graph[u][v])
                dist[v] = dist[u] + graph[u][v];
    }
}

int main() {
    while (scanf("%d", &n) == 1) {
        for (int i = 0; i < MAXN; i++)
            for (int j = 0; j < MAXN; j++)
                graph[i][j] = (i == j) ? 0 : INF;
        for (int i = 2; i <= n; i++) {
            for (int j = 1; j < i; j++) {
                int w; char ch;
                if (scanf("%d", &w) == 1) graph[i][j] = graph[j][i] = w;
                else scanf("%c", &ch); // 跳過非數字
            }
        }
        dijkstra(1);
        int ans = 0;
        for (int i = 1; i <= n; i++) ans = max(ans, dist[i]);
        cout << ans << endl;
    }
    return 0;
}