POJ&&HDU A Walk Through the Forest

                                                              A Walk Through the Forest

题目链接：杭电 北大

题目分析：

   写了一个c++版的，居然在杭电上CE了。在北大上就神奇的过了。

   回归本题题意。题目意思是，给你一些点和一些边，而你选的边需要满足如下这个条件。（A,B）：存在一条从B出发回家的路径，比所有从A出发回家的路径都短。你的任务是计算一共有几条不同的回家路径。注意本题不是叫你求最短路径，而是在解题过程中要运用到最短路径而已。

算法分析：

  Dijkstra+记忆化搜索。

#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;

const int INF = 1e7;
const int maxn = 1e3 + 5;
struct HeapNode{           //堆
   int d,u;
   bool operator<(const HeapNode &rhs)const{
      return d > rhs.d;
   }
};
struct Edge{                 //边集
   int from,to,dist;
};
vector<Edge> edges;
vector<int> G[maxn];        //每个节点出发的编号
int d[maxn],dp[maxn];
void Clear()
{
    for(int i = 0;i < maxn;++i)
      G[i].clear(),d[i] = INF,dp[i] = -1;
}
void AddEdge(int from,int to,int d)
{
    Edge e;
    e.from = from;
    e.to = to;
    e.dist = d;
    edges.push_back(e);
    int m = edges.size();
    G[from].push_back(m-1);
}
void Dijkstra(int s)
{
    bool done[maxn];
    HeapNode tmp;
    priority_queue<HeapNode> Q;
    memset(done,0,sizeof(done));
    d[s] = 0;
    tmp.d = d[s];
    tmp.u = s;
    Q.push(tmp);
    while(!Q.empty())
    {
        HeapNode x = Q.top();
        Q.pop();
        int u = x.u;
        if(done[u])
          continue;
        done[u] = true;
        for(int i = 0;i < (int)G[u].size();++i){
           Edge& e = edges[G[u][i]];
           if(d[e.to] > d[u]+e.dist){
              d[e.to] = d[u]+e.dist;
              //p[e.to] = G[i][i];
              tmp.d = d[e.to];
              tmp.u = e.to;
              Q.push(tmp);
           }
        }
    }
}
int DP(int s)
{
    if(dp[s]!=-1)
      return dp[s];
    if(s == 2)
      return 1;
    int sum = 0;
    for(int i = 0;i < (int)G[s].size();++i){
       Edge& e = edges[G[s][i]];
       if(d[s] > d[e.to]){
          if(dp[e.to] != -1)
            sum += dp[e.to];
          else
            sum += DP(e.to);
       }
    }
    //sum += dp[s];
    dp[s] = sum;
    return dp[s];
}
int main()
{
//    freopen("Input.txt","r",stdin);
    int n,m;
    while(scanf("%d",&n),n)
    {
        int x,y,w;
        Clear();
        scanf("%d",&m);
        for(int i = 0;i < m;++i){
           scanf("%d%d%d",&x,&y,&w);
           AddEdge(x,y,w);
           AddEdge(y,x,w);
        }
        Dijkstra(2);             
        printf("%d\n", DP(1));
    }
    return 0;
}

在北大上才能过。

Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta

#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;

const int INF = 1e7;
const int maxn = 1e3 + 5;
struct HeapNode{
   int d,u;
   bool operator<(const HeapNode &rhs)const{
      return d > rhs.d;
   }
};
struct Edge{
   int from,to,dist;
};
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn],dp[maxn];
void Clear()
{
    for(int i = 0;i < maxn;++i)
      G[i].clear(),d[i] = INF,dp[i] = -1;
}
void AddEdge(int from,int to,int d)
{
    edges.push_back((Edge){from,to,d});
    int m = edges.size();
    G[from].push_back(m-1);
}
void Dijkstra(int s)
{
    bool done[maxn];
    priority_queue<HeapNode> Q;
    memset(done,0,sizeof(done));
    d[s] = 0;
    Q.push((HeapNode){d[s],s});
    while(!Q.empty())
    {
        HeapNode x = Q.top();
        Q.pop();
        int u = x.u;
        if(done[u])
          continue;
        done[u] = true;
        for(int i = 0;i < (int)G[u].size();++i){
           Edge& e = edges[G[u][i]];
           if(d[e.to] > d[u]+e.dist){
              d[e.to] = d[u]+e.dist;
              //p[e.to] = G[i][i];
              Q.push((HeapNode){d[e.to],e.to});
           }
        }
    }
}
int DP(int s)
{
    if(dp[s]!=-1)
      return dp[s];
    if(s == 2)
      return 1;
    int sum = 0;
    for(int i = 0;i < (int)G[s].size();++i){
       Edge& e = edges[G[s][i]];
       if(d[s] > d[e.to]){
          if(dp[e.to] != -1)
            sum += dp[e.to];
          else
            sum += DP(e.to);
       }
    }
    //sum += dp[s];
    dp[s] = sum;
    return dp[s];
}
int main()
{
//    freopen("Input.txt","r",stdin);
    int n,m;
    while(scanf("%d",&n),n)
    {
        int x,y,w;
        Clear();
        scanf("%d",&m);
        for(int i = 0;i < m;++i){
           scanf("%d%d%d",&x,&y,&w);
           AddEdge(x,y,w);
           AddEdge(y,x,w);
        }
        Dijkstra(2);   //for(int i = 1;i <= n;++i)printf("id = %d d = %d\n",i,d[i]);
        printf("%d\n", DP(1));
    }
    return 0;
}

POJ&&HDU A Walk Through the Forest,布布扣,bubuko.com

POJ&&HDU A Walk Through the Forest

原文：http://blog.csdn.net/zhongshijunacm/article/details/20711713