1111. Online Map (30)解题报告

求时间最短路径时，如果时间相同，则选择通过交叉路口最少的路线。仔细审题，否则会容易出错。

#define _CRT_SECURE_NO_WARNINGS#include <cstdio>#include <cstdlib>#include <queue>#include <vector>#include <stack>using namespace std;struct node {int len, time;int v1, v2;};void Dijkstra1(vector<node> arr[], int n, vector<int> path[]);void Dijkstra2(vector<node> arr[], int n, vector<int> path[]);void DFS1(vector<node> arr[], vector<int> path[], int index);void DFS2(vector<node> arr[], vector<int> path[], int index);const int MAX = 100000000;stack<int> road;int time = MAX, len = MAX, source, destination;int main(void){int i, n, m, v1, v2, one_way, t, l, t2, d1;node tmp;vector<node> *arr;vector<int> *path;stack<int> path1, path2;scanf("%d %d", &n, &m);arr = new vector<node>[n];path = new vector<int>[n];for (i = 0; i < m; i++) {scanf("%d %d %d %d %d", &v1, &v2, &one_way, &l, &t);tmp.len = l;tmp.time = t;tmp.v2 = v2;tmp.v1 = v1;arr[v1].push_back(tmp);if(!one_way) {tmp.v2 = v1;tmp.v1 = v2;arr[v2].push_back(tmp);}}scanf("%d %d", &source, &destination);Dijkstra1(arr, n, path);DFS1(arr, path, destination);d1 = len;path1 = road;for (i = 0; i < n; i++) {path[i].clear();}time = MAX;while (!road.empty()) {road.pop();}Dijkstra2(arr, n, path);DFS2(arr, path, destination);t2 = time;path2 = road;if (path1 == path2) {printf("Distance = %d; Time = %d: %d", d1, t2, path2.top());path2.pop();while (!path2.empty()) {printf(" -> %d", path2.top());path2.pop();}putchar('\n');}else {printf("Distance = %d: %d", d1, path1.top());path1.pop();while (!path1.empty()) {printf(" -> %d", path1.top());path1.pop();}putchar('\n');printf("Time = %d: %d", t2, path2.top());path2.pop();while (!path2.empty()) {printf(" -> %d", path2.top());path2.pop();}putchar('\n');}return 0;}void Dijkstra1(vector<node> arr[], int n, vector<int> path[]) {int i, j, *dist = new int[n];bool *visit = (bool *)calloc(n, sizeof(bool));for (i = 0; i < n; i++) {dist[i] = MAX;}dist[source] = 0;int min, index;for (i = 0; i < n; i++) {min = MAX;index = -1;for (j = 0; j < n; j++) {if (!visit[j] && dist[j] < min) {min = dist[j];index = j;}}if (index == -1) {return;}visit[index] = true;for (j = 0; j < arr[index].size(); j++) {if (dist[index] + arr[index][j].len < dist[arr[index][j].v2]) {dist[arr[index][j].v2] = dist[index] + arr[index][j].len;path[arr[index][j].v2].clear();path[arr[index][j].v2].push_back(index);}else if (dist[index] + arr[index][j].len == dist[arr[index][j].v2]) {path[arr[index][j].v2].push_back(index);}}}len = dist[destination];return;}void Dijkstra2(vector<node> arr[], int n, vector<int> path[]){int i, j, *t = new int[n];bool *visit = (bool *)calloc(n, sizeof(bool));for (i = 0; i < n; i++) {t[i] = MAX;}t[source] = 0;int min, index;for (i = 0; i < n; i++) {min = MAX;index = -1;for (j = 0; j < n; j++) {if (!visit[j] && t[j] < min) {min = t[j];index = j;}}if (index == -1) {return;}visit[index] = true;for (j = 0; j < arr[index].size(); j++) {if (t[index] + arr[index][j].time < t[arr[index][j].v2]) {t[arr[index][j].v2] = t[index] + arr[index][j].time;path[arr[index][j].v2].clear();path[arr[index][j].v2].push_back(index);}else if (t[index] + arr[index][j].time == t[arr[index][j].v2]) {path[arr[index][j].v2].push_back(index);}}}time = t[destination];return;}void DFS1(vector<node> arr[],  vector<int> path[], int index) {static stack<int> s;static int t = 0;if (index == source) {s.push(index);if (t < time) {time = t;road = s;}s.pop();}else {int i, j;s.push(index);for (i = 0; i < path[index].size(); i++) {for (j = 0; j < arr[path[index][i]].size(); j++) {if (arr[path[index][i]][j].v2 == index) {t += arr[path[index][i]][j].time;break;}}DFS1(arr, path, path[index][i]);t -= arr[path[index][i]][j].time;}s.pop();}}void DFS2(vector<node> arr[], vector<int> path[], int index){static stack<int> s;if (index == source) {s.push(index);if (road.size() == 0) {road = s;}else if (s.size() < road.size()) {road = s;}s.pop();}else {int i, j;s.push(index);for (i = 0; i < path[index].size(); i++) {DFS2(arr, path, path[index][i]);}s.pop();}}