hdu3665Floyd解法

题目链接：http://icpc.njust.edu.cn/Problem/Hdu/3665/

Floyd是经典的dp算法，将迭代过程分成n个阶段，经过n个阶段的迭代所有点对之间的最短路径都可以求出，时间复杂度是O(n^3)。

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef unsigned int ui;
typedef long long ll;
typedef unsigned long long ull;
#define pf printf
#define mem(a,b) memset(a,b,sizeof(a))
#define prime1 1e9+7
#define prime2 1e9+9
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define scand(x) scanf("%llf",&x) 
#define f(i,a,b) for(int i=a;i<=b;i++)
#define scan(a) scanf("%d",&a)
#define mp(a,b) make_pair((a),(b))
#define P pair<int,int>
#define dbg(args) cout<<#args<<":"<<args<<endl;
#define inf 0x3f3f3f3f
const int maxn=1e3+5;
int n,m,t;
int edge[maxn][maxn],d[maxn][maxn];
int ans=inf;
void init()
{
    f(i,0,n)
        f(j,0,n)
        {
            if(i==j)d[i][j]=0;
            else d[i][j]=inf;
        }
}
void floyd()
{
    f(k,0,n)
        f(i,0,n)
            f(j,0,n)
            {
                d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
            }
}
int main()
{
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    while(scan(n)!=EOF)
    {
        init();
        int mi,pi;
        f(i,0,n-1)
        {
            scan(mi);
            scan(pi);
            if(pi)d[i][n]=0;
            int u,v;
            while(mi--)
            {
                scan(u);
                scan(v);
                d[i][u]=v;
            }
        }
        floyd();
        pf("%d
",d[0][n]);
     } 
 }