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  • 【HDOJ】2385 Stock

    水题,逆向做+优先级队列。

      1 /* 2385 */
      2 #include <iostream>
      3 #include <sstream>
      4 #include <string>
      5 #include <map>
      6 #include <queue>
      7 #include <set>
      8 #include <stack>
      9 #include <vector>
     10 #include <deque>
     11 #include <algorithm>
     12 #include <cstdio>
     13 #include <cmath>
     14 #include <ctime>
     15 #include <cstring>
     16 #include <climits>
     17 #include <cctype>
     18 #include <cassert>
     19 #include <functional>
     20 #include <iterator>
     21 #include <iomanip>
     22 using namespace std;
     23 //#pragma comment(linker,"/STACK:102400000,1024000")
     24 
     25 #define sti                set<int>
     26 #define stpii            set<pair<int, int> >
     27 #define mpii            map<int,int>
     28 #define vi                vector<int>
     29 #define pii                pair<int,int>
     30 #define vpii            vector<pair<int,int> >
     31 #define rep(i, a, n)     for (int i=a;i<n;++i)
     32 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
     33 #define clr                clear
     34 #define pb                 push_back
     35 #define mp                 make_pair
     36 #define fir                first
     37 #define sec                second
     38 #define all(x)             (x).begin(),(x).end()
     39 #define SZ(x)             ((int)(x).size())
     40 #define lson            l, mid, rt<<1
     41 #define rson            mid+1, r, rt<<1|1
     42 
     43 typedef struct node_t {
     44     int p, m;
     45     
     46     friend bool operator< (const node_t& a, const node_t& b) {
     47         return a.p < b.p;
     48     }
     49     
     50 } node_t;
     51 
     52 const int maxn = 1e5+5;
     53 node_t nd[maxn];
     54 int X[maxn];
     55 int n;
     56 
     57 void solve() {
     58     priority_queue<node_t> Q;
     59     node_t d;
     60     int x;
     61     int ans = 0;
     62     
     63     per(i, 0, n) {
     64         Q.push(nd[i]);
     65         x = X[i];
     66         while (!Q.empty() && x>0) {
     67             d = Q.top();
     68             Q.pop();
     69             if (d.m > x) {
     70                 d.m -= x;
     71                 ans += d.p * x;
     72                 x = 0;
     73                 Q.push(d);
     74             } else {
     75                 x -= d.m;
     76                 ans += d.p * d.m;
     77             }
     78         }
     79     }
     80     
     81     printf("%d
    ", ans);
     82 }
     83 
     84 int main() {
     85     ios::sync_with_stdio(false);
     86     #ifndef ONLINE_JUDGE
     87         freopen("data.in", "r", stdin);
     88         freopen("data.out", "w", stdout);
     89     #endif
     90     
     91     int t;
     92     
     93     scanf("%d", &t);
     94     while (t--) {
     95         scanf("%d", &n);
     96         rep(i, 0, n)
     97             scanf("%d %d %d", &X[i], &nd[i].p, &nd[i].m);
     98         solve();    
     99     }
    100     
    101     #ifndef ONLINE_JUDGE
    102         printf("time = %d.
    ", (int)clock());
    103     #endif
    104     
    105     return 0;
    106 }
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  • 原文地址:https://www.cnblogs.com/bombe1013/p/5173991.html
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