Trip 图dp

Trip

Time Limit 1000ms

Memory Limit 65536K

description

　　Xiao wang has a new car, so he wants to have a trip from his home to hefei, however, there is no road from his house directly to Hefei, so he has to go through a number of cities to get to hefei, his car takes one unit of time driving a unit distance at first, but his car will be slow after every city that he pass, so he will take one more unit of time to take a unit of distance after every city (that is, if he has passed k cities, then he needs k units of time to take a unit of distance, we think his home is the first city), can you tell him how to spend the least time to reach Hefei?
							

input

　　The first line contains two numbers n and m (0 < n ≤ 200, 0 < m < n×(n-1)/2 ), n is the number of city, 1 is xiaowang's home, n is hefei, m is the number of road, bi-directional edge, then the following m line contains 3 numbers x, y, z, means there is a road from x to y, the distance is z.(You can make sure that there is always a path exist between his home ande hefei).
							

output

　　The least time xiao wang must use from his home to hefei.
							

sample_input

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

sample_output

7
							

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f[i][j]表示经过i个点从1到达j所用的最短距离。

f[i][k]=min(f[i-1][j]+i*a[j][k])，即经过i个点从1到达k所用的最短距离等于经过i-1个点从j到达k的最小值(1<j<n)。

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

using namespace std;

int n,m;
const int OO=1e9;
long long a[222][222];
long long f[222][222];

int main()
{
    while (~scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                a[i][j]=f[i][j]=OO;
            }
        }
        for (int i=1; i<=m; i++)
        {
            int u,v,c;
            scanf("%d%d%d",&u,&v,&c);
            a[u][v]=a[v][u]=c;
            if(u==1) f[1][v]=c;
            if(v==1) f[1][u]=c;
        }
        long long _min=a[1][n];
        for(int i=2; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(f[i-1][j]!=OO)
                {
                    for(int k=1; k<=n; k++)
                    {
                        if(f[i-1][j]+i*a[j][k]<f[i][k])
                        {
                            f[i][k]=f[i-1][j]+i*a[j][k];
                        }
                    }
                }
            }
        }
        for(int i=1; i<=n-1; i++)
        {
            if(f[i][n]<_min) _min=f[i][n];
        }
        printf("%I64d\n",_min);
    }
    return 0;
}