POJ 3613

//共T条路，求从S到E经过N条路径的最短路径 
#include <iostream>
#include <fstream>
#include <string.h>
#include <algorithm>
using namespace std;

#define CIR(n, m) for (int i=0; i<n; i++) for (int j=0; j<m; j++)
const int INF = 0x3fffffff;//少加个3就wa 

struct Node
{
    Node()
    {
          CIR(301,301) a[i][j]=INF;
    }
    int a[301][301];
    int* operator [](int x)//貌似运算符重载，可惜不懂啊  
    {
          return a[x];
    }
};

int N, T, S, E;
int f[10001] = {0}, top=0;
Node map;

void floyd(Node &a, Node &b, Node &c)
{
    Node temp;
    for (int k=1; k<=top; k++)
        for (int i=1; i<=top; i++)
            for (int j=1; j<=top; j++)
                temp[i][j] = min(temp[i][j], a[i][k]+b[k][j]);
    CIR(301,301) c[i][j] = temp[i][j];
}

int find(int num)
{
    Node ans;
    for (int i=0; i<301; i++) ans[i][i]=0;
    while (num)
    {
        if (num&1) 
          floyd(map, ans, ans);
        floyd(map, map, map);
        num >>= 1;
    }
    return ans[f[S]][f[E]];
}

int main()
{
    int i,j,k;
    int length, from, to;
    cin>>N>>T>>S>>E;
    for (i=0; i<T; i++)
    {
        cin>>length>>from>>to;
        if (!f[from]) f[from]=++top;
        if (!f[to]) f[to]=++top;
        map[f[from]][f[to]] = map[f[to]][f[from]] = length;
    }  
    cout<<find(N)<<endl;
    return 0;
}

不懂啊     不懂啊