【CF】3B Lorry

这道题目网上有几个题解，均有问题。其实就是简单的贪心+排序，没必要做的那么复杂。
一旦tot+curv > v时，显然curv==2, 有三种可能：
（1）取出最小的curv==1的pp，装入当前的p；
（2）取出后续最大的curv==1的p，并且装入；
（3）当前已经是最优的（即后续不存在curv==1的类型），同时前一个的pp比当前的p更优。（这种情况不需要特判）

  1 /* 3B */
  2 #include <iostream>
  3 #include <string>
  4 #include <map>
  5 #include <queue>
  6 #include <set>
  7 #include <stack>
  8 #include <vector>
  9 #include <deque>
 10 #include <algorithm>
 11 #include <cstdio>
 12 #include <cmath>
 13 #include <ctime>
 14 #include <cstring>
 15 #include <climits>
 16 #include <cctype>
 17 #include <cassert>
 18 #include <functional>
 19 #include <iterator>
 20 #include <iomanip>
 21 using namespace std;
 22 //#pragma comment(linker,"/STACK:102400000,1024000")
 23 
 24 #define sti                set<int>
 25 #define stpii            set<pair<int, int> >
 26 #define mpii            map<int,int>
 27 #define vi                vector<int>
 28 #define pii                pair<int,int>
 29 #define vpii            vector<pair<int,int> >
 30 #define rep(i, a, n)     for (int i=a;i<n;++i)
 31 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 32 #define clr                clear
 33 #define pb                 push_back
 34 #define mp                 make_pair
 35 #define fir                first
 36 #define sec                second
 37 #define all(x)             (x).begin(),(x).end()
 38 #define SZ(x)             ((int)(x).size())
 39 #define lson            l, mid, rt<<1
 40 #define rson            mid+1, r, rt<<1|1
 41 
 42 typedef struct node_t {
 43     int t, p, id;
 44     friend bool operator< (const node_t& a, const node_t& b) {
 45         if (a.p == b.p)
 46             return a.t < b.t;
 47         return a.p > b.p;
 48     }
 49 } node_t;
 50 
 51 const int maxn = 1e5+5;
 52 node_t nd[maxn];
 53 
 54 int main() {
 55     ios::sync_with_stdio(false);
 56     #ifndef ONLINE_JUDGE
 57         freopen("data.in", "r", stdin);
 58         freopen("data.out", "w", stdout);
 59     #endif
 60     
 61     int n, v;
 62     
 63     scanf("%d %d", &n, &v);
 64     rep(i, 1, n+1) {
 65         scanf("%d %d", &nd[i].t, &nd[i].p);
 66         nd[i].id = i;
 67         if (nd[i].t == 1)
 68             nd[i].p += nd[i].p;
 69     }
 70     
 71     sort(nd+1, nd+1+n);
 72     
 73     int i, j, k, p, pp = 0, pid, tot = 0;
 74     int ans = 0, tmp;
 75     vi vc;
 76     
 77     i = 1;
 78     while (i <= n) {
 79         if (nd[i].t == 1) {
 80             k = 1;
 81             p = nd[i].p>>1;
 82             pp = p;
 83             pid = nd[i].id;
 84         } else {
 85             k = 2;
 86             p = nd[i].p;
 87         }
 88         
 89         if (tot+k <= v) {
 90             tot += k;
 91             ans += p;
 92             vc.pb(nd[i].id);
 93         } else if (pp) {
 94             // tot+k > v
 95             // must be k == 2
 96             // and pre has a 1
 97             j = i+1;
 98             while (j<=n && nd[j].t!=1)
 99                 ++j;
100             
101             // no type 1 but may be pp < p (becase p is half if the nd[pid].p).
102             if (j > n) {
103                 tmp = 0;
104             } else {
105                 tmp = nd[j].t>>1;
106             }
107             // two way to solve
108             if (p-pp > tmp) {
109                 // erase pid and add new id
110                 for (vi::iterator iter=vc.begin(); iter!=vc.end(); ++iter) {
111                     if (*iter == pid) {
112                         vc.erase(iter);
113                         break;
114                     }
115                 }
116                 vc.pb(nd[i].id);
117                 ans += p-pp;
118             } else if (tmp) {
119                 // add nd[j].id;
120                 vc.pb(nd[j].id);
121                 ans += tmp;
122             }
123             
124             break;
125         }
126         
127         if (tot == v)
128             break;
129         ++i;
130     }
131     
132     printf("%d
", ans);
133     rep(i, 0, SZ(vc)) {
134         printf("%d ", vc[i]);
135     }
136     putchar('
');
137     
138     #ifndef ONLINE_JUDGE
139         printf("time = %d.
", (int)clock());
140     #endif
141     
142     return 0;
143 }