poj 3368 Frequent values

最新推荐文章于 2019-10-17 16:55:14 发布

转载最新推荐文章于 2019-10-17 16:55:14 发布 · 71 阅读

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原文链接：http://www.cnblogs.com/LyonLys/archive/2012/08/18/poj_3368_Lyon.html

本文探讨了解决POJ 3368问题的方法，对比了RMQ（区间最值查询）与线段树两种算法的实现细节及优劣。通过具体代码示例展示了两种方法在解决区间查询问题上的应用。

http://poj.org/problem?id=3368

　　看完RMQ的课件，里面用了这题做例题，所以就试下用RMQ来解这题。

　　RMQ总算是看懂而且会用了。RMQ跟线段树有点相似，也是二分区间来快速求出最值。不过，线段树的适用范围明显更广，RMQ主要是用作离线查询区间最值的。然而居然推荐这题拿来做RMQ，还真让我不解.....不过也没关系，也可以做，就是query的时候显得有点麻烦罢了！

　　query的修改耗了我不少时间，因为RMQ查询的区间长度总是2的n次方，所以如果我的查询结果横跨两个区间，这时的操作就有点麻烦了。这时查询的复杂度升至O(log n)了。这个使用限制还是我一直找不到错误数据看了一下别人的题解才明白啊！也是因为这样，RMQ还是最好在Minimum/Maximum Query的时候用。

好难看的代码（RMQ版1200ms+）：

View Code

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <algorithm>
  5 #include <cmath>
  6 
  7 const int maxn = 100005;
  8 const int maxm = 17;
  9 
 10 struct sec{
 11     int m;
 12     int lv, ln;
 13     int rv, rn;
 14     void add(int a, int b, int c, int d, int e){
 15         m = a;
 16         lv = b; ln = c;
 17         rv = d; rn = e;
 18     }
 19 }s[maxn][maxm];
 20 int ep[maxm + 2];
 21 
 22 void pre(){
 23     ep[0] = 1;
 24     for (int i = 1; i < maxm + 2; i++)
 25         ep[i] = ep[i - 1] << 1;
 26 #ifndef ONLINE_JUDGE
 27     for (int i = 1; i < maxm + 2; i++)
 28         printf("%d\n", ep[i]);
 29 #endif
 30 }
 31 
 32 void scan(int &n){
 33     char ch;
 34 
 35     while (((ch = getchar()) < '0' || ch > '9') && ch != '-');
 36     if (ch != '-'){
 37         n = ch - '0';
 38         while ((ch = getchar()) >= '0' && ch <= '9')
 39             n = n * 10 + ch - '0';
 40     }
 41     else{
 42         n = 0;
 43         while ((ch = getchar()) >= '0' && ch <= '9')
 44             n = n * 10 + '0' - ch;
 45     }
 46 }
 47 
 48 void initRMQ(int n){
 49     int tmp, ln, rn;
 50 
 51     for (int j = 1; ep[j] <= n; j++){
 52         for (int i = 1; i + ep[j] <= n + 1; i++){
 53             if (s[i][j - 1].m > s[i + ep[j - 1]][j - 1].m) tmp = s[i][j - 1].m;
 54             else tmp = s[i + ep[j - 1]][j - 1].m;
 55             if (s[i][j - 1].rv == s[i + ep[j - 1]][j - 1].lv){
 56                 if (tmp < s[i][j - 1].rn + s[i + ep[j - 1]][j - 1].ln)
 57                     tmp = s[i][j - 1].rn + s[i + ep[j - 1]][j - 1].ln;
 58             }
 59             if (s[i][j - 1].lv == s[i + ep[j - 1]][j - 1].lv) ln = ep[j - 1] + s[i + ep[j - 1]][j - 1].ln;
 60             else ln = s[i][j - 1].ln;
 61             if (s[i][j - 1].rv == s[i + ep[j - 1]][j - 1].rv) rn = ep[j - 1] + s[i][j - 1].rn;
 62             else rn = s[i + ep[j - 1]][j - 1].rn;
 63             s[i][j].add(tmp, s[i][j - 1].lv, ln, s[i + ep[j - 1]][j - 1].rv, rn);
 64         }
 65     }
 66 }
 67 
 68 int query(int l, int r){
 69     int len;
 70     int ret, tmp;
 71 
 72     len = (int)floor(log((double)r - l + 1) / log(2.0));
 73 #ifndef ONLINE_JUDGE
 74     printf("len %d\n", len);
 75 #endif
 76     if (s[l][len].m > s[r - ep[len] + 1][len].m) ret = s[l][len].m;
 77     else ret = s[r - ep[len] + 1][len].m;
 78     tmp = s[l][len].rn;
 79     while (s[l][len].rv == s[l + ep[len]][0].lv && l + ep[len] <= r){
 80         l += ep[len];
 81         len = (int)floor(log((double)r - l + 1) / log(2.0));
 82 #ifndef ONLINE_JUDGE
 83         printf("len %d\n", len);
 84 #endif
 85         tmp += s[l][len].ln;
 86         if (s[l][len].lv != s[l][len].rv || l >= r) break;
 87     }
 88 
 89     if (tmp > ret) return tmp;
 90     else return ret;
 91 }
 92 
 93 void RMQ(int n){
 94     int m;
 95     int x, y;
 96 
 97     scan(m);
 98     for (int i = 1; i <= n; i++){
 99         scan(x);
100         //scanf("%d", &x);
101         s[i][0].add(1, x, 1, x, 1);
102     }
103 
104     initRMQ(n);
105     for (int i = 0; i < m; i++){
106         scan(x); scan(y);
107         //scanf("%d%d", &x, &y);
108         printf("%d\n", query(x, y));
109     }
110 }
111 
112 int main(){
113     int n;
114 #ifndef ONLINE_JUDGE
115     freopen("in", "r", stdin);
116 #endif
117     pre();
118     while (true){
119         memset(s, 0, sizeof(s));
120         scan(n);
121         //scanf("%d", &n);
122         if (!n) return 0;
123         RMQ(n);
124     }
125 }

用线段树写：

View Code

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <cmath>
  5 #include <algorithm>
  6 
  7 #define lson l, m, rt << 1
  8 #define rson m + 1, r, (rt << 1) | 1
  9 
 10 const int maxn = 100005;
 11 struct seg{
 12     int m;
 13     int lv;
 14     int ln;
 15     int rv;
 16     int rn;
 17 };
 18 int lv[maxn << 2], ln[maxn << 2], mx[maxn << 2], rv[maxn << 2], rn[maxn << 2];
 19 
 20 void init(int n){
 21     for (int i = 0; i <= n << 2; i++)
 22         lv[i] = ln[i] = rv[i] = rn[i] = mx[i] = 0;
 23 }
 24 
 25 void scan(int &n){
 26     char ch;
 27 
 28     while (((ch = getchar()) < '0' || ch > '9') && ch != '-');
 29     if (ch != '-'){
 30         n = ch - '0';
 31         while ((ch = getchar()) >= '0' && ch <= '9')
 32             n = n * 10 + ch - '0';
 33     }
 34     else{
 35         n = 0;
 36         while ((ch = getchar()) >= '0' && ch <= '9')
 37             n = n * 10 + '0' - ch;
 38     }
 39 }
 40 
 41 void up(int rt){
 42     int l = rt << 1;
 43     int r = (rt << 1) | 1;
 44 
 45     if (mx[l] > mx[r]) mx[rt] = mx[l];
 46     else mx[rt] = mx[r];
 47     if (rv[l] == lv[r]){
 48         if (mx[rt] < rn[l] + ln[r])
 49             mx[rt] = rn[l] + ln[r];
 50     }
 51     lv[rt] = lv[l];
 52     rv[rt] = rv[r];
 53     if (lv[l] == lv[r]) ln[rt] = ln[l] + ln[r];
 54     else ln[rt] = ln[l];
 55     if (rv[r] == rv[l]) rn[rt] = rn[r] + rn[l];
 56     else rn[rt] = rn[r];
 57 }
 58 
 59 void build(int l, int r, int rt){
 60     if (l == r){
 61         int x;
 62 
 63         scan(x);
 64         lv[rt] = rv[rt] = x;
 65         mx[rt] = ln[rt] = rn[rt] = 1;
 66 
 67         return ;
 68     }
 69     int m = (l + r) >> 1;
 70 
 71     build(lson);
 72     build(rson);
 73     up(rt);
 74 }
 75 
 76 seg query(int L, int R, int l, int r, int rt){
 77     seg ret;
 78     if (L <= l && r <= R){
 79         ret.m = mx[rt];
 80         ret.lv = lv[rt];
 81         ret.ln = ln[rt];
 82         ret.rv = rv[rt];
 83         ret.rn = rn[rt];
 84 #ifndef ONLINE_JUDGE
 85         printf("%2d: %2d %2d %2d %2d %2d\n", rt, ret.m, ret.lv, ret.ln, ret.rv, ret.rn);
 86 #endif
 87         return ret;
 88     }
 89 
 90     seg rret;
 91     int m = (l + r) >> 1;
 92 
 93     if (L <= m){
 94         ret = query(L, R, lson);
 95         if (m < R){
 96             rret = query(L, R, rson);
 97             if (ret.m < rret.m) ret.m = rret.m;
 98             if (ret.rv == rret.lv){
 99                 if (ret.m < ret.rn + rret.ln)
100                     ret.m = ret.rn + rret.ln;
101             }
102             if (ret.lv == rret.lv) ret.ln += rret.ln;
103             if (ret.rv == rret.rv) ret.rn += rret.rn;
104             else ret.rn = rret.rn; // 开始的时候漏了这句，wa了好久..- -
105             ret.rv = rret.rv;
106         }
107     }
108     else if (m < R) ret = query(L, R, rson);
109 #ifndef ONLINE_JUDGE
110     printf("%2d: %2d %2d %2d %2d %2d\n", rt, ret.m, ret.lv, ret.ln, ret.rv, ret.rn);
111 #endif
112 
113     return ret;
114 }
115 
116 void deal(int n){
117     int m;
118     int x, y;
119 
120     init(n);
121     scan(m);
122     build(1, n, 1);
123 #ifndef ONLINE_JUDGE
124     for (int i = 0; i < (n << 2); i++)
125         printf("%2d: %2d %2d %2d %2d %2d\n", i, mx[i], lv[i], ln[i], rv[i], rn[i]);
126     puts("");
127 #endif
128     for (int i = 0; i < m; i++){
129         scan(x); scan(y);
130         printf("%d\n", query(x, y, 1, n, 1).m);
131     }
132 }
133 
134 int main(){
135     int n;
136 #ifndef ONLINE_JUDGE
137     freopen("in", "r", stdin);
138 #endif
139     while (true){
140         scan(n);
141         if (!n) return 0;
142         deal(n);
143     }
144 }

　　貌似写出来的代码也是要挺长的，而且这题的合并要注意，我就因为更新的时候漏了一丁点东西，测试了不下100组数据，测了我一个晚上才找出问题的根源....囧！以后要加倍注意线段树更新的严谨啊！

——written by Lyon

转载于:https://www.cnblogs.com/LyonLys/archive/2012/08/18/poj_3368_Lyon.html