转载请注明出处： http://www.cnblogs.com/fraud/ ——by fraud
Hiking
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 724 Accepted Submission(s): 384
Special Judge
Problem Description
There are n soda conveniently labeled by 1,2,…,n. beta, their best friends, wants to invite some soda to go hiking. The i-th soda will go hiking if the total number of soda that go hiking except him is no less than li and no larger than ri. beta will follow the rules below to invite soda one by one:
1. he selects a soda not invited before;
2. he tells soda the number of soda who agree to go hiking by now;
3. soda will agree or disagree according to the number he hears.
Note: beta will always tell the truth and soda will agree if and only if the number he hears is no less than li and no larger than ri, otherwise he will disagree. Once soda agrees to go hiking he will not regret even if the final total number fails to meet some soda‘s will.
Help beta design an invitation order that the number of soda who agree to go hiking is maximum.
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first contains an integer n (1≤n≤105), the number of soda. The second line constains n integers l1,l2,…,ln. The third line constains n integers r1,r2,…,rn. (0≤li≤ri≤n)
It is guaranteed that the total number of soda in the input doesn‘t exceed 1000000. The number of test cases in the input doesn‘t exceed 600.
Output
For each test case, output the maximum number of soda. Then in the second line output a permutation of 1,2,…,n denoting the invitation order. If there are multiple solutions, print any of them.
Sample Input
4
8
4 1 3 2 2 1 0 3
5 3 6 4 2 1 7 6
8
3 3 2 0 5 0 3 6
4 5 2 7 7 6 7 6
8
2 2 3 3 3 0 0 2
7 4 3 6 3 2 2 5
8
5 6 5 3 3 1 2 4
6 7 7 6 5 4 3 5
Sample Output
7
1 7 6 5 2 4 3 8
8
4 6 3 1 2 5 8 7
7
3 6 7 1 5 2 8 4
0
1 2 3 4 5 6 7 8
水题，先按li排序，然后不断地塞进优先队列
1 /**
2 * code generated by JHelper
3 * More info: https://github.com/AlexeyDmitriev/JHelper
4 * @author xyiyy @https://github.com/xyiyy
5 */
6
7 #include <iostream>
8 #include <fstream>
9
10 //#####################
11 //Author:fraud
12 //Blog: http://www.cnblogs.com/fraud/
13 //#####################
14 //#pragma comment(linker, "/STACK:102400000,102400000")
15 #include <iostream>
16 #include <sstream>
17 #include <ios>
18 #include <iomanip>
19 #include <functional>
20 #include <algorithm>
21 #include <vector>
22 #include <string>
23 #include <list>
24 #include <queue>
25 #include <deque>
26 #include <stack>
27 #include <set>
28 #include <map>
29 #include <cstdio>
30 #include <cstdlib>
31 #include <cmath>
32 #include <cstring>
33 #include <climits>
34 #include <cctype>
35
36 using namespace std;
37 #define mp(X, Y) make_pair(X,Y)
38 #define pb(X) push_back(X)
39 #define rep(X, N) for(int X=0;X<N;X++)
40 #define rep2(X, L, R) for(int X=L;X<=R;X++)
41 typedef pair<int, int> PII;
42
43 typedef pair<PII, int> PIII;
44 priority_queue<PIII, vector<PIII>, greater<PIII> > q;
45 int f[100010];
46 int l[100010];
47 int r[100010];
48 vector<int> v[100010];
49
50 class hdu5360 {
51 public:
52 void solve(std::istream &in, std::ostream &out) {
53 int n;
54 in >> n;
55 for (int i = 1; i <= n; i++)in >> l[i];
56 for (int i = 1; i <= n; i++) {
57 in >> r[i];
58 if (l[i])v[l[i]].pb(i);
59 else q.push(mp(mp(r[i], l[i]), i));
60 }
61 int num = n;
62 int ans = 0;
63 int sz = 0;
64 while (n) {
65 if (q.empty()) {
66 rep2(i, ans + 1, n) {
67 rep(j, v[i].size()) {
68 f[sz++] = v[i][j];
69 }
70 v[i].clear();
71 }
72 break;
73 }
74 PIII p = q.top();
75 q.pop();
76 int y = p.first.first;
77 int x = p.first.second;
78 int z = p.second;
79 f[sz++] = z;
80 if (x <= ans && y >= ans) {
81 ans++;
82 rep(i, v[ans].size()) {
83 int j = v[ans][i];
84 q.push(mp(mp(r[j], l[j]), j));
85 }
86 v[ans].clear();
87 }
88 }
89 out << ans << endl;
90 rep(i, sz) {
91 if (i)out << " ";
92 out << f[i];
93 }
94 out << endl;
95 }
96 };
97
98 int main() {
99 std::ios::sync_with_stdio(false);
100 std::cin.tie(0);
101 hdu5360 solver;
102 std::istream &in(std::cin);
103 std::ostream &out(std::cout);
104 int n;
105 in >> n;
106 for (int i = 0; i < n; ++i) {
107 solver.solve(in, out);
108 }
109
110 return 0;
111 }