POJ 3159 Candies （差分约束#1）

来源：互联网发布：飞行器发动机知乎编辑：程序博客网时间：2024/05/01 22:49
用了两种方法：用栈的SPFA和用优先队列的Dijkstra都是500ms+
 #include <cstdio>#include <algorithm>#include <queue> #include <cstdlib> #include <cstring> #include <stack> #define N 30005#define M 150005using namespace std; const int inf = (-1u >> 1); int n, m;typedef pair<int, int> pii; int first[N], d[N]; int next[M], u[M], v[M], w[M];bool inq[N];int SPFA(int s, int n)//用栈 {       for (int i = 1; i <= n; i++)d[i] =  inf;       d[s] = 0;       stack<int> q;       memset(inq, false, sizeof(inq));        q.push(s);        while (!q.empty())       {              int x = q.top();q.pop();              inq[x] = false;              for (int e = first[x]; e != -1; e = next[e])if (d[v[e]] > d[u[e]] + w[e])              {                    d[v[e]] = d[u[e]] + w[e];                    if (!inq[v[e]])                    {                              inq[v[e]] = true;                              q.push(v[e]);                    }              }       }       return d[n]; }void dijkstra(int s, int n){        for (int i = 1;i <= n; i++)d[i] =  inf;        d[s] = 0;        priority_queue<pii, vector<pii>, greater<pii> > q;        q.push(make_pair(d[s], s));        while (!q.empty())        {              pii u = q.top();q.pop();              int x = u.second;              if (u.first != d[x])continue;              for (int e = first[x]; e != -1; e = next[e])if (d[v[e]] > d[x] + w[e])              {                     d[v[e]] = d[x] + w[e];                     q.push(make_pair(d[v[e]], v[e]));              }        }} int main(){     // FILE* fp = fopen("in.txt", "r");       while (scanf( "%d", &n) != EOF && n)      {            scanf( "%d", &m);            int a, b, c;            int e = 0;             for (int i = 1;i <= n ; i++)first[i] = -1;            for (int i = 0; i < m; i++)            {                   scanf( "%d %d %d", &a, &b, &c);                  next[e] = first[a];                  first[a] = e;                  w[e] = c;                  u[e] = a, v[e] = b;                  e++;            }            dijkstra(1, n);            printf("%d\n",  d[n]);       }      return 0;}