1111 Online Map（30分）

题目翻译：

就是求解一个点到另一个点的最短路径。

不过由于限制条件很多，还得分开求两次，所以写起来就很繁。

题解思路：

可以用dijkstra或者dfs，用后者的话最后一个测试点可能会超时。

代码：

dfs（最后一个测试点超时）：

#include<bits/stdc++.h>
using namespace std;
int N, M;//结点数量，边的数量
vector<int> v[501];
int dis[501][501];
int cost[501][501];
int mindis[501];
int min_len = INT_MAX, min_cost = INT_MAX;
vector<int> path1, path2, temp;
int so, des;
int temp_cost = INT_MAX;void dfs(int curnode, int curlen,int curcost)
{if (curlen > mindis[curnode]) return;elsemindis[curnode] = curlen;temp.push_back(curnode);if (curnode == des){if (curlen < min_len){min_len = curlen;path1 = temp;temp_cost = curcost;}if (curlen == min_len&&curcost<temp_cost)path1 = temp;}else {for (auto i : v[curnode])dfs(i, curlen + dis[curnode][i], curcost + cost[curnode][i]);}temp.pop_back();
}void dfs2(int curnode, int curlen)
{if (curlen > mindis[curnode]) return;elsemindis[curnode] = curlen;temp.push_back(curnode);if (curnode == des){if (curlen < min_cost){min_cost = curlen;path2 = temp;}if (curlen == min_cost && temp.size() < path2.size())path2 = temp;}else {for (auto i : v[curnode])dfs2(i, curlen + cost[curnode][i]);}temp.pop_back();
}int main()
{cin >> N >> M;int q, w, e, r, t;for (int i = 0;i < M;i++){cin >> q >> w >> e >> r >> t;if (e)//单向{v[q].push_back(w);dis[q][w] = r;cost[q][w] = t;}else//双向{v[q].push_back(w);v[w].push_back(q);dis[q][w] = dis[w][q] = r;cost[q][w] = cost[w][q] = t;}}for (int i = 0;i < 501;i++)mindis[i] = INT_MAX;cin >> so >> des;dfs(so, 0, 0);for (int i = 0;i < 501;i++)mindis[i] = INT_MAX;temp.clear();dfs2(so, 0);if (path1 != path2){cout << "Distance = " << min_len << ":";for (int i = 0;i < path1.size();i++){if (i)cout << " ->";cout << " " << path1[i];}cout << endl;}elsecout << "Distance = " << min_len << "; ";cout << "Time = " << min_cost << ":";for (int i = 0;i < path2.size();i++){if (i)cout << " ->";cout << " " << path2[i];}
}

dijkstra（AC，转载于AC代码）

#include<bits/stdc++.h>
using namespace std;int n, m, source, dest, min_distance, min_time;
vector<int> ans_t, ans_d, dpre(500, -1), tpre(500, -1);
vector<vector<int>> times(500, vector<int>(500, -1)), distances(500, vector<int>(500, -1)), graph(500,vector<int>(500, 0));
void dfs1(int x){if(x == -1) return;dfs1(dpre[x]);ans_d.push_back(x);
}
void dfs2(int x){if(x == -1) return;dfs2(tpre[x]);ans_t.push_back(x);
}
void dijkstra1(int beg){vector<int> lowcost(500, 0x3f3f3f3f), vis(500, 0), dtime(500, 0x3f3f3f3f);lowcost[beg] = 0;dpre[beg] = -1;for (int i = 0; i < n;i++){int Min = 0x3f3f3f3f, k = -1;for (int j = 0; j < n;j++){if(vis[j] == 0 && lowcost[j] < Min){Min = lowcost[j];k = j;}}if(k == -1) break;vis[k] = 1;for (int j = 0; j < n;j++){if(vis[j] == 0 && graph[j][k] == 1){//如果满足条件则更新信息，这里要注意逻辑短路现象if(lowcost[j] > lowcost[k] + distances[j][k] || (lowcost[j] == lowcost[k] + distances[j][k] && dtime[k] + times[k][j] < dtime[j])){dpre[j] = k;dtime[j] = dtime[k] + times[k][j];lowcost[j] = lowcost[k] + distances[j][k];}}}}min_distance = lowcost[dest];dfs1(dest);
}void dijkstra2(int beg){vector<int> lowcost(500, 0x3f3f3f3f), vis(500, 0), intersections(500, 0x3f3f3f3f);lowcost[beg] = 0;tpre[beg] = -1;for (int i = 0; i < n;i++){int Min = 0x3f3f3f3f, k = -1;for (int j = 0; j < n;j++){if(vis[j] == 0 && lowcost[j] < Min){Min = lowcost[j];k = j;}}if(k == -1) break;vis[k] = 1;for (int j = 0; j < n;j++){if(vis[j] == 0 && graph[j][k] == 1){if(lowcost[j] > lowcost[k] + times[j][k] || (lowcost[j] == lowcost[k] + times[j][k] && intersections[k] + 1 < intersections[j])){tpre[j] = k;intersections[j] = intersections[k] + 1;lowcost[j] = lowcost[k] + times[j][k];}}}}min_time = lowcost[dest];dfs2(dest);
}int main(){int s, e, a, l, t, isSame = 1;cin >> n >> m;for (int i = 0; i < m;i++){cin >> s >> e >> a >> l >> t;graph[s][e] = graph[e][s] = 1;times[s][e] = times[e][s] = t;distances[s][e] = distances[e][s] = l;}cin >> source >> dest;dijkstra1(source);dijkstra2(source);if(ans_d.size() == ans_t.size()){for (int i = 0; i < ans_t.size();i++){if(ans_t[i] != ans_d[i]){isSame = 0;break;//能省点时间就省点}}}elseisSame = 0;if(isSame == 0){printf("Distance = %d: %d", min_distance, ans_d[0]);for (int i = 1; i < ans_d.size();i++)printf(" -> %d", ans_d[i]);printf("\nTime = %d: %d", min_time, ans_t[0]);for (int i = 1; i < ans_t.size();i++)printf(" -> %d", ans_t[i]);}else{printf("Distance = %d; Time = %d: %d", min_distance, min_time, ans_t[0]);for (int i = 1; i < ans_t.size();i++)printf(" -> %d", ans_t[i]);}cout << endl;return 0;
}