最短路径问题

 

Dijstra算法求最短路径

九度1008

题目描述：

给你n个点，m条无向边，每条边都有长度d和花费p，给你起点s终点t，要求输出起点到终点的最短距离及其花费，如果最短距离有多条路线，则输出花费最少的。

输入：

输入n,m，点的编号是1~n,然后是m行，每行4个数 a,b,d,p，表示a和b之间有一条边，且其长度为d，花费为p。最后一行是两个数 s,t;起点s，终点t。n和m为0时输入结束。
(1

输出：

输出 一行有两个数， 最短距离及其花费。

样例输入：

3 2
1 2 5 6
2 3 4 5
1 3
0 0

样例输出：

9 11

#include
#include
#include
using namespace std;
struct E {
	int next;
	int dis;
	int price;
};
vector edge[1000];
int main() {
	int n, m;
	int i, j;
	while (scanf("%d%d", &n, &m) != EOF) {
		if (n == 0 && m == 0) break;
		for (i = 1; i <= n; i++) {
			edge[i].clear();
		}
		E temp;
		while (m--) {
			int a, b, d, p;
			scanf("%d%d%d%d", &a, &b, &d, &p);
			temp.dis = d;
			temp.price = p;
			temp.next = b;
			edge[a].push_back(temp);
			temp.next = a;
			edge[b].push_back(temp);
		}
		int s, t;
		scanf("%d%d", &s, &t);
		bool mark[1000];
		int dis[1000];
		int pri[1000];
		int newp;
		for (i = 1; i <= n; i++) {
			mark[i] = false;
			dis[i] = -1;
			pri[i] = 0;
		}
		mark[s] = true;
		dis[s] = 0;
		newp = s;
		for (i = 1; i < n; i++) {
			for (j = 0; j < edge[newp].size(); j++) {
				int tn = edge[newp][j].next;
				int td = edge[newp][j].dis;
				int tp = edge[newp][j].price;
				if (mark[tn] == true) continue;
				if (dis[tn] == -1 || dis[newp] + td < dis[tn] || (dis[newp] + td == dis[tn] && pri[newp] + tp < pri[tn])) {
					dis[tn] = dis[newp] + td;
					pri[tn] = pri[newp] + tp;
				}
			}
			int min = 123123123;
			for (j = 1; j <= n; j++) {
				if (mark[j] == true) continue;
				if (dis[j] == -1) continue;
				if (dis[j] < min) {
					min = dis[j];
					newp = j;
				}
			}
			mark[newp] = true;
		}
		printf("%d %d", dis[t], pri[t]);
	}
}

邻接矩阵表示

#include
#include
using namespace std;
int n, m;
bool mark[1001];
int s, e;
int dis[1001];
int edge[1001][1001];
int main() {
	int i, j;
	while (scanf("%d%d", &n, &m) != EOF) {
		if (n == 0 && m == 0) break;
		for (i = 1; i <= m; i++) {
			int a, b, c;
			scanf("%d%d%d", &a, &b, &c);
			edge[a][b] = edge[b][a] = c;
		}
		//初始化
		for (i = 0; i < 1001; i++) {
			dis[i] = 123123123;
			mark[i] = false;
		}
		s = 1, e = n;
		int min = 123123123;
		int newp=1;
		dis[1] = 0;
		mark[1] = true;
		for (i = 1; i <= n; i++) {
			for (j = 1; j <= n; j++) {
				if (mark[j] == false && min > dis[j]) {
					min = dis[j];
					newp = j;
				}
			}
			mark[newp] = true;
			for (j = 1; j <= n; j++) {
				if (dis[j] > dis[newp] + edge[newp][j]) {
					dis[j] = dis[newp] + edge[newp][j];
				}
			}
		}
		printf("%d", dis[n]);
	}
}