hdu1142 dijstra + 记忆化搜索

来源：互联网发布：视频捕捉软件编辑：程序博客网时间：2024/06/06 03:53

题目难以理解，不过最后百度后才理解的大意，其中有一个关键的意思是，如果A到B直接相连，且从A到终点的最短路径大于从B到终点的最短路径，则该人可以从A走向B，问这样走下去，合法路径一共有多少。

直接贴代码，该题让我练习了一下dijkstra和记忆化搜索，说实话，记忆化搜索对于我来说，还真是硬伤。还得没事多练习下。

/*His office is numbered 1,and his house is numbered 2.The first  N,1 < N ≤ 1000,  M.The followingM1 ≤ d ≤ 1000000There is at most one path between any pair of intersections.Outputassume that this number does not exceed 2147483647Sample Input5 61 3 21 4 23 4 31 5 124 2 345 2 247 81 3 11 4 13 7 17 4 17 5 16 7 15 2 16 2 10Sample Output24*/#include <cstdio>#include <iostream>#include <vector>#include <queue>#include <utility> //pair头文件#include <cstring>#include <string>using namespace std;const int V_MAX = 1005;typedef struct edge{    int to ,cost;    edge(int a, int b){        to = a;        cost = b;    }}edge;unsigned int dp[V_MAX];int N,M;vector<edge> G[V_MAX];typedef pair<int, int > P;const unsigned int INF = 1<<32 -1;void cal_minpath(int s){    for(int i=1; i<=N; i++)dp[i] = INF;    dp[s] = 0;    priority_queue<P, vector<P >,greater<P > > pque;    pque.push(P(0,s));    while(!pque.empty())    {        P temp = pque.top();        pque.pop();        int len = temp.first;        int V = temp.second;        if(len > dp[V])continue;        for(int i=0;i<G[V].size(); i++)        {            int fir = G[V][i].to;            int sec = G[V][i].cost;            if(dp[fir] > dp[V]+sec)            {                dp[fir] = dp[V] + sec;                pque.push(P(dp[fir], fir));            }        }    }}int visited[V_MAX];unsigned int sum[V_MAX];unsigned int dfs(int s){    if(sum[s]){return sum[s];}    if(s == 2){return 1;}    unsigned int psum = 0;    for(int i=0; i<G[s].size(); i++)    {        edge e = G[s][i];        if(dp[s] > dp[e.to])        {            if(sum[e.to])psum += sum[e.to];            else psum += dfs(e.to);        }    }    sum[s] = psum;    return sum[s];}int main(){    while(true)    {        int from,to,cost;        memset(visited,0,sizeof(visited));        memset(sum, 0, sizeof(sum));        memset(G,0,sizeof(G));        scanf("%d", &N);        if(N == 0)break;        scanf("%d", &M);        for(int i=0; i<M; i++)        {            scanf("%d%d%d",&from, &to, &cost);            G[from].push_back(edge(to,cost));            G[to].push_back(edge(from,cost));        }        cal_minpath(2);        cout << dfs(1) << endl;    }    return 0;}

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