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传送门:https://pintia.cn/problem-sets/9 ... /994805464397627392
解:
迪杰斯特拉 + DFS (模板)
- #include <cstdio>
- #include <climits>
- #include <vector>
- #include <algorithm>
- using namespace std;
- const int maxn = 510;
- int n, m, c1, c2;
- int G[maxn][maxn];
- int cost[maxn][maxn];
- int d[maxn];//存放最短路径长度
- bool vis[maxn];//是否访问过
- vector<int> pre[maxn];//存放顶点在最短路径中的前驱
- void DJS(int s)
- {
- fill(d, d + n, INT_MAX);
- d[s] = 0;
- for (int i = 0; i < n; i++)
- {
- int u = -1, min = INT_MAX;
- for (int i = 0; i < n; i++)//查找未被访问过且距离最短的顶点
- if (!vis[i] && d[i] < min) min = d[u = i];
- if (u == -1) return;//如果剩下的顶点与源点不通,返回
- vis[u] = true;
- for (int v = 0; v < n; v++)
- {
- if (vis[v] || G[u][v] == INT_MAX) continue;
- int t = d[u] + G[u][v];
- if (t < d[v])
- {
- d[v] = t;
- pre[v].clear();
- pre[v].push_back(u);
- }
- else if (t == d[v])
- pre[v].push_back(u);
- }
- }
- }
- int opt_cost = INT_MAX;//存放最少花费
- vector<int> path, temp_path;
- void DFS(int v)
- {
- if (v == c1)//递归边界,如果遍历到起点
- {
- temp_path.push_back(v);
- int val = 0;
- for (int i = temp_path.size() - 1; i > 0; i--)
- val += cost[temp_path[i]][temp_path[i - 1]];
- if (val < opt_cost)
- {
- opt_cost = val;
- path = temp_path;
- }
- temp_path.pop_back();
- return;
- }
- temp_path.push_back(v);
- for (int i = 0; i < pre[v].size(); i++)
- DFS(pre[v][i]);
- temp_path.pop_back();
- }
- int main(void)
- {
- scanf("%d %d %d %d", &n, &m, &c1, &c2);
-
- fill(G[0], G[0] + maxn * maxn, INT_MAX);
-
- while (m--)
- {
- int a, b, l, c;
- scanf("%d %d %d %d", &a, &b, &l, &c);
- G[a][b] = G[b][a] = l;
- cost[a][b] = cost[b][a] = c;
- }
- DJS(c1);
- DFS(c2);
- for (int i = path.size() - 1; i >= 0; i--)
- printf("%d ", path[i]);
- printf("%d %d", d[c2], opt_cost);
- return 0;
- }
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