【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）

/* 3B */
#include <iostream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000")

#define sti                set<int>
#define stpii            set<pair<int, int> >
#define mpii            map<int,int>
#define vi                vector<int>
#define pii                pair<int,int>
#define vpii            vector<pair<int,int> >
#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define clr                clear
#define pb                 push_back
#define mp                 make_pair
#define fir                first
#define sec                second
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1

typedef struct node_t {
    int t, p, id;
    friend bool operator< (const node_t& a, const node_t& b) {
        if (a.p == b.p)
            return a.t < b.t;
        return a.p > b.p;
    }
} node_t;

const int maxn = 1e5+5;
node_t nd[maxn];

int main() {
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
        freopen("data.out", "w", stdout);
    #endif
    
    int n, v;
    
    scanf("%d %d", &n, &v);
    rep(i, 1, n+1) {
        scanf("%d %d", &nd[i].t, &nd[i].p);
        nd[i].id = i;
        if (nd[i].t == 1)
            nd[i].p += nd[i].p;
    }
    
    sort(nd+1, nd+1+n);
    
    int i, j, k, p, pp = 0, pid, tot = 0;
    int ans = 0, tmp;
    vi vc;
    
    i = 1;
    while (i <= n) {
        if (nd[i].t == 1) {
            k = 1;
            p = nd[i].p>>1;
            pp = p;
            pid = nd[i].id;
        } else {
            k = 2;
            p = nd[i].p;
        }
        
        if (tot+k <= v) {
            tot += k;
            ans += p;
            vc.pb(nd[i].id);
        } else if (pp) {
            // tot+k > v
            // must be k == 2
            // and pre has a 1
            j = i+1;
            while (j<=n && nd[j].t!=1)
                ++j;
            
            // no type 1 but may be pp < p (becase p is half if the nd[pid].p).
            if (j > n) {
                tmp = 0;
            } else {
                tmp = nd[j].t>>1;
            }
            // two way to solve
            if (p-pp > tmp) {
                // erase pid and add new id
                for (vi::iterator iter=vc.begin(); iter!=vc.end(); ++iter) {
                    if (*iter == pid) {
                        vc.erase(iter);
                        break;
                    }
                }
                vc.pb(nd[i].id);
                ans += p-pp;
            } else if (tmp) {
                // add nd[j].id;
                vc.pb(nd[j].id);
                ans += tmp;
            }
            
            break;
        }
        
        if (tot == v)
            break;
        ++i;
    }
    
    printf("%d
", ans);
    rep(i, 0, SZ(vc)) {
        printf("%d ", vc[i]);
    }
    putchar('
');
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.
", (int)clock());
    #endif
    
    return 0;
}