Big Truck

/problems/bigtruck/file/statement/en/img-0001.jpg

Your boss has hired you to drive a big truck, transporting items between two locations in a city. You’re given a description of the city, with locations of interest and the lengths of roads between them. Your boss requires that you take a shortest path between the starting and ending location, and she’ll check your odometer when you’re done to make sure you didn’t take any unnecessary side trips. However, your friends know you have plenty of unused space in the truck, and they have asked you to stop by several locations in town, to pick up items for them. You’re happy to do this for them. You may not be able to visit every location to pick up everything your friends want, but you’d like to pick up as many items as possible on your trip, as long as it doesn’t make the path any longer than necessary.

$\includegraphics[width=.7\textwidth ]{graph}$

Figure 1: Illustrations of the first two sample inputs

The two graphs above show examples of what the city may look like, with nodes representing locations, edges representing roads and dots inside the nodes representing items your friends have asked you to pick up. Driving through a location allows you to pick up all the items there; it’s a big truck, with no limit on the items it can carry. In the graph on the left, for example, you have to drive the big truck from location $1$

Input

The first line of input contains an integer, $n$

Output

If it’s not possible to travel from location $1$

Sample Input 1	Sample Output 1
6 1 1 2 3 1 0 7 1 2 2 2 3 3 3 6 4 1 4 4 4 3 2 4 5 3 5 6 2	9 5

Sample Input 2	Sample Output 2
9 1 1 1 1 1 1 1 1 1 10 1 2 3 2 5 3 1 6 2 6 7 2 7 5 2 5 3 1 3 4 2 4 9 3 5 8 2 8 9 4	12 7

Sample Input 3	Sample Output 3
2 5 5 0	impossible

题解：

其实就是一个最短路，只不过过程需要记录一下每个点的权值，并保留，

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#include <bits/stdc++.h>
using namespace std;
const int MAXN=110;
const int INF=0x3f3f3f3f;
int n;
int val[MAXN];
int mapp[MAXN][MAXN];
int dis[MAXN],ans[MAXN];
bool vis[MAXN];
void dij()
{
    for(int i=1;i<=n;i++)
        dis[i]=mapp[1][i];
    dis[1]=0;
    int MAIN,V=-1,k;
    for (int i = 1; i <=n ; ++i) {
        MAIN=INF;
        k=V;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&MAIN>dis[j])
            {
                MAIN=dis[j];
                V=j;
            }

        }
        vis[V]=true;
        for (int j = 1; j <=n ; ++j) {
            if(!vis[j]&&dis[j]>dis[V]+mapp[V][j])
            {
                dis[j]=dis[V]+ mapp[V][j];
                ans[j]=ans[V]+val[j];
            } else if(!vis[j]&&dis[j]==dis[V]+mapp[V][j])
            {
                ans[j]=max(ans[j],ans[V]+val[j]);
            }
        }
    }

    if(dis[n]==0)
        printf("impossible\n");
    else
    printf("%d %d\n",dis[n],ans[n]+val[1]);
}
int main()
{
    scanf("%d",&n);
    memset(mapp,0x3f, sizeof(mapp));
    for (int i =1; i <=n ; ++i) {
        scanf("%d",&val[i]);
        mapp[i][i]=0;
    }
    int m;scanf("%d",&m);
    int x,y,z;

    while(m--)
    {
        scanf("%d%d%d",&x,&y,&z);
        mapp[x][y]=mapp[y][x]=z;
    }
    dij();
    return 0;

}
//HDU-3790

/*

 9
1 1 1 1 1 1 1 1 1
10
1 2 3
2 5 3
1 6 2
6 7 2
7 5 2
5 3 1
3 4 2
4 9 3
5 8 2
8 9 4

 6
1 1 2 3 1 0
7
1 2 2
2 3 2
3 6 4
1 4 4
4 3 2
4 5 3
5 6 2

 */

Input

Output

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