Dijkstra算法详解与应用-优快云博客

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1010;
const int inf = 0x3f3f3f3f;
int map1[maxn][maxn];
int dis[maxn];
bool vis[maxn];
int n,m;
void dijkstra() {
	int i,j; 
	memset(vis,false,sizeof(vis));
	memset(dis,inf,sizeof(dis));
	dis[1] = 0;
	vis[1] = 1;
	for(i = 1; i<=n; i++) {
		dis[i] = map1[1][i];
	}
	for(int time = 1;time<=n-1;time++){
		int minn = inf;
		int p = 1;//start
		for(i = 1;i<=n;i++){
			if(dis[i]<minn&&vis[i]!=1){
				minn = dis[i];
				p = i;
			}
		}
		vis[p] = 1;
		for(i = 1;i<=n;i++){
			dis[i] = min(dis[i],dis[p]+map1[p][i]);
		}
	}
}

以上为模板

										过冬
						 		Tags:     弗洛伊德  Dijk
						Time Limit:  1 s      Memory Limit:   32 MB

Description
天气逐渐变冷，年老体弱的越越鸟打算去南方某座温度适宜的城市过冬，但由于翅膀严重老化，最远飞行距离有限，请你为可怜的越越鸟计算飞行所需的最短距离，以让它能做好心理准备。

Input
输入包含多组测试数据。

每组输入第一行为两个正整数n（n<=20）和m（m<=n*(n-1)/2），n表示城市个数，m表示线段个数。（线段为两个城市间的连接线）

接下来m行，每行输入三个整数a，b，和l（l<=10^9），表示a市与b市之间存在一条线段，线段长度为l。（a与b不同）

每组最后一行输入两个整数x和y，表示问题：x为越越鸟现在所在的城市，y为越越鸟打算飞往过冬的城市。城市标号为1~n。

Output
对于每组输入，输出x市与y市之间的最短距离，如果x市与y市之间非连通，则输出“No path”。

Samples
input:
4 4
1 2 4
1 3 1
1 4 1
2 3 1
2 4
output:
3

#include<bits/stdc++.h>
using namespace std;
const int maxn = 22;
const int inf = 0x3f3f3f3f;
int map1[maxn][maxn];
bool vis[maxn];
int dis[maxn];
int m,n;
void dijkstra(int start){
	int i,j;
	memset(vis,false,sizeof(vis));
	memset(dis,inf,sizeof(dis));
	dis[start] = 0;
	vis[start] = 1;
	for(i = 1;i<=n;i++){
		dis[i] = map1[start][i];
	}
	for(int time = 1;time<=n-1;time++){
		int minn = inf;
		int p = start;
		for(i = 1;i<=n;i++){
			if(dis[i]<minn&&vis[i]!=1){
				minn = dis[i];
				p = i;
			}
		}
		vis[p] = 1;
		for(i = 1;i<=n;i++){
			dis[i] = min(dis[i],dis[p]+map1[p][i]);
		}
	}
}
int main(){
	int i,j,a,b,l,x,y;
	while(scanf("%d %d",&n,&m)!=EOF){
		memset(map1,inf,sizeof(map1));
		for(i = 1;i<=n;i++){
			map1[i][i] = 0;
		}
		for(i = 0;i<m;i++){
			scanf("%d %d %d",&a,&b,&l);
			map1[a][b] = map1[b][a] = l;
		}
		scanf("%d %d",&x,&y);
		dijkstra(x);
		if(dis[y]==inf){
			printf("No path\n");
		}
		else
			printf("%d\n",dis[y]);
	}
}

模板代入题解

								Silver Cow Party

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1…N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow’s return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input
Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2… M+1: Line i+1 describes road i with three space-separated integers: Ai, Bi, and Ti. The described road runs from farm Ai to farm Bi, requiring Ti time units to traverse.
Output
Line 1: One integer: the maximum of time any one cow must walk.
Sample Input
4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3
Sample Output
10

Hint
Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn = 1010;
const int inf = 0x3f3f3f3f;
bool vis[maxn];
int map1[maxn][maxn];
int dis[maxn];
int sum[maxn];
int n,m,x;
void dijkstra(int start){
	int i,j;
	memset(vis,false,sizeof(vis));
	memset(dis,inf,sizeof(dis));
	dis[start] = 0;
	vis[start] = 1;
	for(i = 1;i<=n;i++){
		dis[i] = map1[start][i];
	}
	for(int time = 1;time<=n-1;time++){
		int minn = inf;
		int p = start;
		for(i = 1;i<=n;i++){
			if(dis[i]<minn&&vis[i]!=1){
				minn = dis[i];
				p = i;
			}
		}
		vis[p] = 1;
		for(i = 1;i<=n;i++){
			dis[i] = min(dis[i],dis[p]+map1[p][i]);
		}
	}
}
int main(){
	int a,b,l,i,maxx = -2,j,temp;
	scanf("%d %d %d",&n,&m,&x);
	memset(map1,inf,sizeof(map1));
	memset(sum,0,sizeof(sum));
	for(i = 1;i<=n;i++){
		map1[i][i] = 0;
	}
	while(m--){
		scanf("%d %d %d",&a,&b,&l);
		map1[a][b] = l;
	}
	dijkstra(x);
	for(i = 1;i<=n;i++){
		if(i!=x&&dis[i]!=inf){
			sum[i] += dis[i];
		}
	}
//	for(i = 1;i<=n;i++){
//		printf("%d ",sum[i]);
//	}
//	puts("");
	for(i = 1;i<=n;i++){
		for(j = i+1;j<=n;j++){
			temp = map1[i][j];
			map1[i][j] = map1[j][i];
			map1[j][i] = temp;
		}
	}
	dijkstra(x);
	maxx = 0;
	for(i = 1;i<=n;i++){
		if(i!=x&&dis[i]!=inf){
			sum[i] += dis[i];
			//printf("%d ",dis[i]);
			maxx = max(maxx,sum[i]);
		}
	}
	//puts("");
	//for(i = 1;i<=n;i++){
	//	printf("%d ",sum[i]);
	//}
	//puts("");
	printf("%d\n",maxx);
}

计算回来的路径矩阵翻转 AC代码