【BZOJ4397】[Usaco2015 dec]Breed Counting【前缀和】【或莫队】【或线段树】【或可持久化线段树】

来源：互联网发布：三菱plc编程入门pdf 编辑：程序博客网时间：2024/05/24 07:09

【题目链接】

数据结构学多了，看到题解发现3个前缀和就搞定了。弱智+2

另外也可以线段树，也可以3个主席树。。。

莫队：

/* Telekinetic Forest Guard */#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 100005, maxsqrtn = 316;int n, m, num[maxn], cnt[5];struct data {int l, r, id;bool operator < (const data &x) const {int a = l / maxsqrtn, b = x.l / maxsqrtn;return a != b ? a < b : r < x.r;}} que[maxn], ans[maxn];inline int iread() {int f = 1, x = 0; char ch = getchar();for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';return f * x;}int main() {n = iread(); m = iread();for(int i = 1; i <= n; i++) num[i] = iread();for(int i = 1; i <= m; i++) que[i].l = iread(), que[i].r = iread(), que[i].id = i;sort(que + 1, que + 1 + m);int l = 1, r = 0;for(int i = 1; i <= m; i++) {while(que[i].l < l) cnt[num[--l]]++;while(r < que[i].r) cnt[num[++r]]++;while(l < que[i].l) cnt[num[l++]]--;while(que[i].r < r) cnt[num[r--]]--;ans[que[i].id] = (data){cnt[1], cnt[2], cnt[3]};}for(int i = 1; i <= m; i++) printf("%d %d %d\n", ans[i].l, ans[i].r, ans[i].id);return 0;}

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