題意：給出n個點，m條邊的圖，求1到n的 最短路。

程式碼：

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <iostream>

using namespace std;

const int M = 1e5 + 10;
const int N = 1e2 + 10;
const int INF = 0x3f3f3f3f;

struct Node {
  int
 
 p, dis;

  Node() {}

  Node(int a, int b) {
    p = a;
    dis = b;
  }

  friend bool operator < (Node a, Node b) {
    return a.dis > b.dis;
  }
};

int x[M], y[M], z[M];
int edge[N][N];
int d[N];
int ct[N];
bool inq[N];
int n, m;

int dij() {
  priority_queue<Node> q;
  q.push(Node(1
 
, 0));
  d[1] = 0;
  while (!q.empty()) {
    Node f = q.top();
    q.pop();
    for (int i = 1; i <= n; i++) {
      if (d[i] > edge[f.p][i] + f.dis) {
        d[i] = edge[f.p][i] + f.dis;
        q.push(Node(i, d[i]));
      }
    }
  }
  return d[n];
}

int bf() {
  d[1] = 0;
  for (int i = 0; i < n; i++) {
    for
 
 (int j = 0; j < 2 * m; j++) {
      int a = x[j], b = y[j], c = z[j];
      d[a] = min(d[a], d[b] + c);
    }
  }
  return d[n];
}

int floyd() {
  for (int i = 1; i <= n; i++)
    edge[i][i] = 0;
  for (int k = 1; k <= n; k++) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        edge[i][j] = min(edge[i][j], edge[i][k] + edge[k][j]);
      }
    }
  }
  return edge[1][n];
}

int spfa() {
  queue<int> q;
  q.push(1);
  d[1] = 0;
  inq[1] = true;
  ct[1] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    inq[u] = false;
    for (int i = 1; i <= n; i++) {
      if (d[i] > d[u] + edge[u][i]) {
        d[i] = d[u] + edge[u][i];
        if (!inq[i]) {
          q.push(i);
          inq[i] = true;
          ct[i]++;
          if (ct[i] == n)
            return -1;
        }
      }
    }
  }
  return d[n];
}

int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    if (!n && !m)
      break;
    memset(edge, INF, sizeof(edge));
    memset(d, INF, sizeof(d));
    memset(inq, false, sizeof(inq));
    memset(ct, 0, sizeof(ct));
    for (int i = 0; i < m; i++) {
      int u, v, w;
      scanf("%d%d%d", &u, &v, &w);
      edge[u][v] = min(edge[u][v], w);
      edge[v][u] = min(edge[v][u], w);
      x[2 * i] = u, y[2 * i] = v, z[2 * i] = w;
      x[2 * i + 1] = v, y[2 * i + 1] = u, z[2 * i + 1] = w;
    }
    // printf("%d\n", dij());
    // printf("%d\n", bf());
    // printf("%d\n", spfa());
    printf("%d\n", floyd());
  }
  return 0;
}

題意：給出n個點的圖，邊以鄰接矩陣的形式給出，q組查詢，每次查詢從u到v的最大安全係數。邊的安全係數之間為相乘的形式。

程式碼：

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <iostream>

using namespace std;

const int N = 1e3 + 10;
const double EPS = 1e-5;

struct Node {
  int p;
  double safe;

  Node() {}

  Node(int a, double b) {
    p = a;
    safe = b;
  }

  friend bool operator < (Node a, Node b) {
    return a.safe < b.safe;
  }
};

double edge[N][N];
double d[N][N];
bool vis[N];
int n, q;

void dij(int s) {
  priority_queue<Node> q;
  q.push(Node(s, 1));
  d[s][s] = 1;
  while (!q.empty()) {
    Node now = q.top();
    q.pop();
    for (int i = 1; i <= n; i++) {
      if (d[s][i] < now.safe * edge[now.p][i]) {
        d[s][i] = now.safe * edge[now.p][i];
        q.push(Node(i, d[s][i]));
      }
    }
  }
}

int main() {
  while (scanf("%d", &n) != EOF) {
    memset(vis, false, sizeof(vis));
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        scanf("%lf", &edge[i][j]);
        d[i][j] = 0;
      }
    }
    scanf("%d", &q);
    for (int i = 1; i <= q; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      if (!vis[u]) {
        vis[u] = true;
        dij(u);
      }
      if (d[u][v] != 0)
        printf("%.3lf\n", d[u][v]);
      else
        printf("What a pity!\n");
    }
  }
  return 0;
}