LA 3938

时间：2014-07-21 00:32:54 阅读：243 评论：0 收藏：0 [点我收藏+]

After doing Ray a great favor to collect sticks for Ray, Poor Neal becomes very hungry. In return for Neal‘s help, Ray makes a great dinner for Neal. When it is time for dinner, Ray arranges all the dishes he makes in a single line (actually this line is very long ... <tex2html_verbatim_mark>, the dishes are represented by 1, 2, 3 ... <tex2html_verbatim_mark>). ``You make me work hard and don‘t pay me! You refuse to teach me Latin Dance! Now it is time for you to serve me", Neal says to himself.

Every dish has its own value represented by an integer whose absolute value is less than 1,000,000,000. Before having dinner, Neal is wondering about the total value of the dishes he will eat. So he raises many questions about the values of dishes he would have.

For each question Neal asks, he will first write down an interval [a, b] <tex2html_verbatim_mark>(inclusive) to represent all the dishes a, a + 1,..., b <tex2html_verbatim_mark>, where a <tex2html_verbatim_mark>and b <tex2html_verbatim_mark>are positive integers, and then asks Ray which sequence of consecutive dishes in the interval has the most total value. Now Ray needs your help.

Input

The input file contains multiple test cases. For each test case, there are two integers n <tex2html_verbatim_mark>and m <tex2html_verbatim_mark>in the first line (n, m < 500000) <tex2html_verbatim_mark>. n <tex2html_verbatim_mark>is the number of dishes and m <tex2html_verbatim_mark>is the number of questions Neal asks.

Then n <tex2html_verbatim_mark>numbers come in the second line, which are the values of the dishes from left to right. Next m <tex2html_verbatim_mark>lines are the questions and each line contains two numbers a <tex2html_verbatim_mark>, b <tex2html_verbatim_mark>as described above. Proceed to the end of the input file.

Output

For each test case, output m <tex2html_verbatim_mark>lines. Each line contains two numbers, indicating the beginning position and end position of the sequence. If there are multiple solutions, output the one with the smallest beginning position. If there are still multiple solutions then, just output the one with the smallest end position. Please output the result as in the Sample Output.

Sample Input

3 1 
1 2 3 
1 1

Sample Output

Case 1: 
1 1

线段树单点更新，虽然思路很简单，但我写的丑爆了！！

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <vector>
  6 #include <assert.h>
  7 #include <queue>
  8 
  9 using namespace std;
 10 
 11 #define read() freopen("sw.in", "r", stdin)
 12 #define ls l, m, rt << 1
 13 #define rs m + 1, r, rt << 1 | 1
 14 
 15 const int MAX = 500007;
 16 const int INF = 1e9 + 7;
 17 typedef long long ll;
 18 int N, M;
 19 ll max_sub[4 * MAX], max_suffix[4 * MAX], max_prefix[4 * MAX];
 20 int id_suffix[4 * MAX], id_prefix[4 * MAX];
 21 int s[4 * MAX], e[4 * MAX];
 22 ll sum[4 * MAX];
 23 struct node {
 24         ll maxsub, maxprefix, maxsuffix , sum;
 25         int idprefix, idsuffix, ids, ide;
 26 };
 27 queue <node> q;
 28 
 29 void push_up(int rt) {
 30         sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
 31 
 32         ll t[3], id = 0;
 33         t[0] = max_sub[rt << 1];
 34         t[1] = max_suffix[rt << 1] + max_prefix[rt << 1 | 1];
 35         t[2] = max_sub[rt << 1 | 1];
 36         for (int i = 0; i <= 2; ++i) {
 37                 //printf("%d\n", t[i]);
 38                 if (t[id] < t[i]) {
 39                         id = i;
 40 
 41                 }
 42         }
 43 
 44         //printf("id = %d\n", id);
 45         max_sub[rt] = t[id];
 46         if (id == 0) {
 47                 s[rt] = s[rt << 1];
 48                 e[rt] = e[rt << 1];
 49         } else if (id == 1) {
 50                 s[rt] = id_suffix[rt << 1];
 51                 e[rt] = id_prefix[rt << 1 | 1];
 52         } else {
 53                 s[rt] = s[rt << 1 | 1];
 54                 e[rt] = e[rt << 1 | 1];
 55         }
 56 
 57         if (max_prefix[rt << 1] < sum[rt << 1] + max_prefix[rt << 1 | 1]) {
 58                 max_prefix[rt] = sum[rt << 1] + max_prefix[rt << 1 | 1];
 59                 id_prefix[rt] = id_prefix[rt << 1 | 1];
 60         } else {
 61                 max_prefix[rt] = max_prefix[rt << 1] ;
 62                 id_prefix[rt] = id_prefix[rt << 1];
 63         }
 64 
 65         if (max_suffix[rt << 1 | 1] <= sum[rt << 1 | 1] + max_suffix[rt << 1]) {
 66                 max_suffix[rt] = sum[rt << 1 | 1] + max_suffix[rt << 1];
 67                 id_suffix[rt] = id_suffix[rt << 1];
 68         } else {
 69                 max_suffix[rt] = max_suffix[rt << 1 | 1];
 70                 id_suffix[rt] = id_suffix[rt << 1 | 1];
 71         }
 72 
 73 
 74 
 75 }
 76 
 77 void build(int l, int r, int rt) {
 78         if (l == r) {
 79                 cin >> sum[rt];
 80                 max_sub[rt] = max_suffix[rt] = max_prefix[rt] = sum[rt];
 81                 id_suffix[rt] = id_prefix[rt] = s[rt] = e[rt] = l;
 82                 return;
 83         }
 84 
 85         int m = (l + r) >> 1;
 86         build(ls);
 87         build(rs);
 88 
 89         push_up(rt);
 90 
 91 }
 92 
 93 void query(int ql, int qr, int l, int r, int rt) {
 94         if (ql <= l && r <= qr) {
 95                 if (q.size() == 1) {
 96                         node x = q.front(), a;
 97                         ll t[3];
 98                         t[0] = x.maxsub;
 99                         t[1] = x.maxsuffix + max_prefix[rt];
100                         t[2] = max_sub[rt];
101                         int id = 0;
102                         for (int i = 0; i <= 2; ++i) {
103                                 if (t[id] < t[i]) {
104                                         id = i;
105                                 }
106                         }
107 
108                         a.maxsub = t[id];
109                         if (id == 0) {
110                                 a.ids = x.ids;
111                                 a.ide = x.ide;
112                         } else if (id == 1) {
113                                 a.ids = x.idsuffix;
114                                 a.ide = id_prefix[rt];
115                         } else {
116                                 a.ids = s[rt];
117                                 a.ide = e[rt];
118                         }
119 
120 
121                         if (max_prefix[rt] <= x.sum + max_prefix[rt]) {
122                                 a.maxprefix = x.sum + max_prefix[rt];
123                                 a.idprefix = id_prefix[rt];
124                         } else {
125                                 a.maxprefix = x.maxprefix;
126                                 a.idprefix = x.idprefix;
127                         }
128 
129                         if (max_suffix[rt] <= sum[rt] + x.maxsuffix) {
130                                 a.maxsuffix = sum[rt] + x.maxsuffix;
131                                 a.idsuffix = x.idsuffix;
132                         } else {
133                                 a.maxsuffix = max_suffix[rt];
134                                 a.idsuffix = id_suffix[rt];
135                         }
136 
137                         q.pop();
138                         q.push(a);
139                     }  else {
140                             node a;
141                             a.ids = s[rt]; a.ide = e[rt];
142                             a.idprefix = id_prefix[rt]; a.idsuffix = id_suffix[rt];
143                             a.maxprefix = max_prefix[rt]; a.maxsub = max_sub[rt];
144                             a.maxsuffix = max_suffix[rt]; a.sum = sum[rt];
145                             q.push(a);
146                     }
147 
148 
149         } else {
150                 int m = (l + r) >> 1;
151                 if (ql <= m) query(ql ,qr, ls);
152                 if (qr > m) query(ql, qr, rs);
153         }
154 }
155 
156 int main()
157 {
158     //read();
159     int ca = 1;
160     while (~scanf("%d%d", &N, &M)) {
161             memset(sum, 0, sizeof(sum));
162             build(1, N, 1);
163             printf("Case %d:\n", ca++);
164             for (int i = 1; i <= M; ++i) {
165                     int l, r;
166                     scanf("%d%d", &l, &r);
167                     query(l, r, 1, N, 1);
168                     node x = q.front(); q.pop();
169                     printf("%d %d\n", x.ids, x.ide);
170 
171 
172             }
173     }
174 
175     //cout << "Hello world!" << endl;
176     return 0;
177 }

View Code

LA 3938,布布扣,bubuko.com

LA 3938

标签：des style blog http color os

原文地址：http://www.cnblogs.com/hyxsolitude/p/3856521.html

踩

(0)

评论一句话评论（0）

分享档案

更多>

2021年07月29日 (22)
2021年07月28日 (40)
2021年07月27日 (32)
2021年07月26日 (79)
2021年07月23日 (29)
2021年07月22日 (30)
2021年07月21日 (42)
2021年07月20日 (16)
2021年07月19日 (90)
2021年07月16日 (35)

周排行