POJ 2387 Bellman双重边

Til the Cows Come Home
Time Limit: 1000MS        Memory Limit: 65536K
Total Submissions: 44129        Accepted: 15001

Description
Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input
* Line 1: Two integers: T and N

* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output
* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100

Sample Output

90

Hint
INPUT DETAILS:

There are five landmarks.

OUTPUT DETAILS:

Bessie can get home by following trails 4, 3, 2, and 1.

关键是多重边，即是双向的，长度不等，用Bellman算法处理比较方便

如：

5 5
2 4 20
2 3 30
3 4 20
4 2 10
1 5 100

dis[4] > dis[2] + 2-4,更新

且dis[2] > dis[4] + 4-2 更新

1点到2点的距离也可以从4-2来松弛

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int maxn = 4000;
const int INF = 99999999;
int main(){
    int dis[maxn],bak[maxn],i,j,n,m,u[2*maxn],v[2*maxn],w[2*maxn],check,flag;
    scanf("%d%d",&m,&n);

    for(i=1;i<=m;i++)
        scanf("%d%d%d",&u[i],&v[i],&w[i]);

    for(i=1;i<=n;i++)
        dis[i]=INF;
        dis[1]=0;

    for(j=1;j<=n-1;j++){
        check=0;
        for(i=1;i<=m;i++){
            if(dis[v[i]] > dis[u[i]] +w[i] ){
                dis[v[i]] = dis[u[i]]+ w[i];
                check = 1;
            } //对所有的单边遍历都不能缩小，那最多边的松弛更不能松弛了
            if(dis[u[i]]>dis[v[i]]+w[i]){
                dis[u[i]] = dis[v[i]]+w[i];
                check = 1;
                }
            }

   if(check==0)
    break;
    }
   printf("%d",dis[n]);
   return 0;

    }

再附上别人的Dijkstra算法做的：

#include <iostream>
using namespace std;
#define inf 1<<29
#define MAXV 1005

int map[MAXV][MAXV];
int n,m;

void dijkstra(){
	int i,j,min,v;
	int d[MAXV];
	bool vis[MAXV];

	for(i=1;i<=n;i++){
		vis[i]=0;
		d[i]=map[1][i];
	}

	for(i=1;i<=n;i++){
		min=inf;
		for(j=1;j<=n;j++)
			if(!vis[j] && d[j]<min){
				v=j;
				min=d[j];
			}
		vis[v]=1;

		for(j=1;j<=n;j++)
			if(!vis[j] && d[j]>map[v][j]+d[v])
				d[j]=map[v][j]+d[v];
	}
	printf("%d
",d[n]);
}

int main(){
	int i,j,a,b,c;
	while(~scanf("%d%d",&m,&n)){
		for(i=1;i<=n;i++)
			for(j=1;j<=n;j++)
				if(i==j)
					map[i][i]=0;
				else map[i][j]=map[j][i]=inf;
		
		for(i=1;i<=m;i++){
			scanf("%d%d%d",&a,&b,&c);
			if(map[a][b]>c) map[a][b]=map[b][a]=c; //关键处理，存下最小的边，DJ算法把有向图转换成无向图处理
		}
		dijkstra();
	}
	return 0;
}