有了一張自駕旅遊路線圖,你會知道城市間的高速公路長度、以及該公路要收取的過路費。現在需要你寫一個程序,幫助前來諮詢的遊客找一條出發地和目的地之間的最短路徑。如果有若干條路徑都是最短的,那麼需要輸出最便宜的一條路徑。
輸入格式:
輸入說明:輸入數據的第1行給出4個正整數NNN、MMM、SSS、DDD,其中NNN(2≤N≤5002\le N\le 5002≤N≤500)是城市的個數,順便假設城市的編號爲0~(N−1N-1N−1);MMM是高速公路的條數;SSS是出發地的城市編號;DDD是目的地的城市編號。隨後的MMM行中,每行給出一條高速公路的信息,分別是:城市1、城市2、高速公路長度、收費額,中間用空格分開,數字均爲整數且不超過500。輸入保證解的存在。
輸出格式:
在一行裏輸出路徑的長度和收費總額,數字間以空格分隔,輸出結尾不能有多餘空格。
輸入樣例:
4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20
輸出樣例:
3 40
參考代碼:
#include <stdio.h>
#include <stdlib.h>
#define MaxNum 500
#define INFINITY 65533
typedef int Vertex;
typedef int Weight;
/*Dijkstral算法變形*/
struct GNode{
int Nv;
int Ne;
Weight G1[MaxNum][MaxNum];
Weight G2[MaxNum][MaxNum];//兩個權重
};
typedef struct GNode *Graph;
struct ENode{
Vertex V1;
Vertex V2;
Weight W1;//距離
Weight W2;//收費
};
typedef struct ENode *Edge;
Vertex FindMin(Graph G, int dist[], int check[])
{
Vertex MinV;
int Mindist = INFINITY;
for (int i = 0; i < G->Nv; i++){
if (check[i] != 1 && dist[i] < Mindist){
Mindist = dist[i];
MinV = i;
}
}
if (Mindist < INFINITY)
return MinV;
else
return -1;
}
void Dijkstra(Graph G, Vertex S, int dist[], int cost[])
//起點S,終點D
{
Vertex V;
int check[MaxNum] = {0,};
for (int i = 0; i < G->Nv; i++){
dist[i] = G->G1[S][i];
cost[i] = G->G2[S][i];
}
check[S] = 1;
while (1){
V = FindMin(G, dist, check);
if (V == -1)
break;
check[V] = 1;
for (int i = 0; i < G->Nv; i++){
if (check[i] == 0 && G->G1[V][i] < INFINITY){
if (dist[V] + G->G1[V][i] < dist[i]){
dist[i] = dist[V] + G->G1[V][i];
cost[i] = cost[V] + G->G2[V][i];
}
else if (dist[V] + G->G1[V][i] == dist[i])
if (cost[V] + G->G2[V][i] < cost[i])
cost[i] = cost[V] + G->G2[V][i];
}
}
}
}
int main(int argc, char const *argv[])
{
Graph G = (Graph)malloc(sizeof(struct GNode));
Edge E = (Edge)malloc(sizeof(struct ENode));
int N, M;
Vertex S, D;//S-->起點 D-->終點
scanf("%d %d %d %d\n", &N, &M, &S, &D);
G->Nv = N;
G->Ne = M;
//初始化圖
for (int i = 0; i < G->Nv; i++ )
for (int j = 0; j < G->Nv; j++){
G->G1[i][j] = INFINITY;
G->G2[i][j] = INFINITY;
}
//無向網圖插入邊
for (int i = 0; i < G->Ne; i++){
scanf("%d %d %d %d", &E->V1, &E->V2, &E->W1, &E->W2);
G->G1[E->V1][E->V2] = E->W1;
G->G1[E->V2][E->V1] = E->W1;
G->G2[E->V1][E->V2] = E->W2;
G->G2[E->V2][E->V1] = E->W2;
}
//Dijstral算法(指定終點)
int dist[MaxNum], cost[MaxNum];
Dijkstra(G, S, dist, cost);
printf("%d %d", dist[D], cost[D]);
system("pause");
return 0;
}