題目描述
給你n個點,m條無向邊,每條邊都有長度d和花費p,給你起點s終點t,要求輸出起點到終點的最短距離及其花費,如果最短距離有多條路線,則輸出花費最少的。
輸入描述:
輸入n,m,點的編號是1~n,然後是m行,每行4個數 a,b,d,p,表示a和b之間有一條邊,且其長度爲d,花費爲p。最後一行是兩個數 s,t;起點s,終點t。n和m爲0時輸入結束。
(1<n<=1000, 0<m<100000, s != t)
輸出描述:
輸出 一行有兩個數, 最短距離及其花費。
示例1
輸入
3 2
1 2 5 6
2 3 4 5
1 3
0 0
輸出
9 11
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
/* 圖的信息 */
typedef struct Edge {
int s;
int e;
int l;
int c;
Edge(int s, int e, int l, int c) {
this->s = s;
this->e = e;
this->l = l;
this->c = c;
}
void print() {
printf("start:%d end:%d length:%d cost:%d \n",this->s,this->e,this->l,this->c);
}
} Edge;
vector<Edge> graph[1001];
typedef struct Point {
int num; // 點的編號
int distanceFromStart; // 從源點的距離
bool operator < (const Point& a) const {
return distanceFromStart > a.distanceFromStart;
}
Point(int n, int d) {
this->num = n;
this->distanceFromStart = d;
}
} Point;
int n, m; // 1 ~ n 編號 ; m個邊
int u, v; // 起點 終點
int INF = 9999999;
int dis[1001];
int cost[1001];
void print() {
cout << "graph" << endl;
for (int i = 1; i <= n; i ++) {
cout << "index:" << i << endl;
for (int j = 0; j <= graph[i].size() - 1; j++) {
graph[i][j].print();
}
}
}
void dijkstra(int u) {
priority_queue<Point> q;
dis[u] = 0;
cost[u] = 0;
q.push(Point(u,0));
while(!q.empty()) {
int father = q.top().num;
// cout << "father:" << father << endl;
q.pop();
for (int i = 0; i <= graph[father].size() - 1; i++) {
int s = graph[father][i].s;
int e = graph[father][i].e;
int l = graph[father][i].l;
int c = graph[father][i].c;
if (dis[e] > dis[s] + l || (dis[e] == dis[s] + l && cost[e] > cost[s] + c)) {
dis[e] = dis[s] + l;
cost[e] = cost[s] + c;
q.push(Point(e,dis[e]));
}
}
}
return ;
}
int main() {
while (cin >> n >> m && n != 0 && m != 0) {
// 初始化
for (int i = 1; i <= n; i++) {
graph[i].clear();
}
for (int i = 1; i <= n; i ++) {
dis[i] = INF;
cost[i] = 0;
}
// 輸入
int s,e,l,c;
for (int i = 1; i <= m; i++) {
cin >> s >> e >> l >> c;
// cout << s << e << l << c << endl;
graph[s].push_back(Edge(s, e, l, c));
graph[e].push_back(Edge(e, s, l, c));
}
// print(); // 測試了 沒問題
//填充dis cost
cin >> u >> v;
dijkstra(u);
cout << dis[v] << " " << cost[v] << endl;
}
}