find the mincost route
2017-07-24 09:47
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find the mincost route杭州有N个景区,景区之间有一些双向的路来连接,现在8600想找一条旅游路线,这个路线从A点出发并且最后回到A点,假设经过的路线为V1,V2,....VK,V1,那么必须满足K>2,就是说至除了出发点以外至少要经过2个其他不同的景区,而且不能重复经过同一个景区。现在8600需要你帮他找一条这样的路线,并且花费越少越好。 Input第一行是2个整数N和M(N <= 100, M <= 1000),代表景区的个数和道路的条数。 接下来的M行里,每行包括3个整数a,b,c.代表a和b之间有一条通路,并且需要花费c元(c <= 100)。Output对于每个测试实例,如果能找到这样一条路线的话,输出花费的最小值。如果找不到的话,输出"It's impossible.".Sample Input
3 3 1 2 1 2 3 1 1 3 1 3 3 1 2 1 1 2 3 2 3 1Sample Output
3 It's impossible.
最小环与Floyd算法讲解:
转载:http://blog.sina.com.cn/s/blog_476a25110100mag6.html
#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int a[100][100],b[100][100];const int maxn=100000000;int main(){int i,j,m,n,k;while(scanf("%d %d",&n,&m)!=EOF){for(i=1;i<=n;i++)for(j=1;j<=n;j++){a[i][j]=maxn;b[i][j]=maxn;}for(i=0;i<m;i++){int a1,a2,s;scanf("%d %d %d",&a1,&a2,&s);if(a[a1][a2]>s)a[a1][a2]=a[a2][a1]=b[a1][a2]=b[a2][a1]=s;}int minn=maxn;for(int k=1;k<=n;k++){for(int i=1;i<k;i++)for(int j=i+1;j<k;j++)minn=min(minn,b[i][j]+a[i][k]+a[k][j]);for(int i=1;i<n;i++)for(int j=1;j<=n;j++)b[i][j]=min(b[i][k]+b[k][j],b[i][j]);}if(minn>=maxn)printf("It's impossible.\n");elseprintf("%d\n",minn);}return 0;}
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