HDU 1142 A Walk Through the Forest(图论Dijkstra)

题意：给你n个点，m条线，问从1到2 有多少种路径数，要求从a点到b点时，b点到2的距离比a到2的距离近。

思路：

1、找出所有点到2的最短路，，

2、用动态规划的方法，从远到近找出到达i点的路径数。存于ans[i];

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>

using namespace std;
const int N = 1009;
const int M = N*N*2;
const int INF = 0x3f3f3f3f;
int n,m;
struct LT{
    int nex,to,dis;
} L[M];
int F[N],cnt;
void add(int f,int t,int d)
{
    L[cnt].dis = d;
    L[cnt].nex = F[f];
    L[cnt].to = t;
    F[f] = cnt++;
}
void init()
{
    int f,t,d;
    memset(F,0,sizeof(F));
    cnt = 1;
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d",&f,&t,&d);
        add(f,t,d);add(t,f,d);
    }
}
int redis[N];///到终点的最短路
struct nod{
    int to,dis;
    bool operator<(const nod t) const
    {
        return dis>t.dis;
    }
};
priority_queue<nod> que;

void remindis()
{
    memset(redis,INF,sizeof(redis));
    while(!que.empty()) que.pop();
    nod e,t;e.dis = 0,e.to = 2;
    que.push(e);
    while(!que.empty())///dijkstra
    {
        e = que.top();que.pop();
        if(redis[e.to]!=INF) continue;
        redis[e.to] = e.dis;
        for(int i = F[e.to];i;i = L[i].nex)
        {
            int to = L[i].to;
            if(redis[to]!=INF) continue;
            t.dis = e.dis+L[i].dis;
            t.to = to;
            que.push(t);
        }
    }
}
struct tnod{
    int to,dis;
    bool operator<(const tnod t) const
    {
        return dis<t.dis;
    }
};
priority_queue<tnod> tque;
int ans[N];
bool visit[N];
void solve()
{
    remindis();
   // for(int i=1;i<=n;i++) cout<<redis[i]<<" ";cout<<endl;
    while(!tque.empty()) tque.pop();
    tnod e,t;e.dis=redis[1],e.to = 1;
    tque.push(e);
    memset(ans,0,sizeof(ans));
    memset(visit,false,sizeof(visit));
    visit[1] = true;ans[1]= 1;
    while(!tque.empty())
    {
        e = tque.top();tque.pop();
       // cout<<e.to<<" "<<e.dis<<endl;
        for(int i=F[e.to];i;i=L[i].nex)
        {
            int to =L[i].to;
            if(redis[to]>=e.dis) continue;
            if(visit[to])
            {
                ans[to]+=ans[e.to];
                continue;
            }
            visit[to] = true;
            ans[to]+=ans[e.to];
            t.dis = redis[to];
            t.to = to;
            tque.push(t);
        }
    }
    printf("%d\n",ans[2]);
}
int main()
{
    freopen("in.txt","r",stdin);
    while(~scanf("%d%d",&n,&m)&&n)
    {
        init();
        solve();
    }
    return 0;
}