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线段树-区间合并uva1400(LA3938)

2014-02-20 21:38 288 查看
"Ray, Pass me the dishes!"

求最大连续和。

写起来太蛋疼了,还是因各种细节过不了,忘了考虑,多组满足时x尽量小,y尽量小,查询的时候也是缺这个少那个。。。

下面是标程,写的很漂亮,多学习!

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn = 500000 + 10;
const int maxnode = 1000000 + 10;
typedef long long LL;
typedef pair<int,int> Interval;

LL prefix_sum[maxn];

LL sum(int L, int R) {
  return prefix_sum[R] - prefix_sum[L-1];
}

LL sum(Interval p) {
  return sum(p.first, p.second);
}

Interval better(Interval a, Interval b) {
  if(sum(a) != sum(b)) return sum(a) > sum(b) ? a : b;
  return a < b ? a : b; // 利用pair自带的字典序
}

int qL, qR;

struct IntervalTree {
  int max_prefix[maxnode];
  int max_suffix[maxnode];
  Interval max_sub[maxnode];

  void build(int o, int L, int R) {
    if(L == R) {
      max_prefix[o] = max_suffix[o] = L;
      max_sub[o] = make_pair(L, L);
    } else {
      int M = L + (R-L)/2;
      // 递归创建子树
      int lc = o*2, rc = o*2+1;
      build(lc, L, M);
      build(rc, M+1, R);

      // 递推max_prefix
      LL v1 = sum(L, max_prefix[lc]);
      LL v2 = sum(L, max_prefix[rc]);
      if(v1 == v2) max_prefix[o] = min(max_prefix[lc], max_prefix[rc]);
      else max_prefix[o] = v1 > v2 ? max_prefix[lc] : max_prefix[rc];

      // 递推max_suffix
      v1 = sum(max_suffix[lc], R);
      v2 = sum(max_suffix[rc], R);
      if(v1 == v2) max_suffix[o] = min(max_suffix[lc], max_suffix[rc]);
      else max_suffix[o] = v1 > v2 ? max_suffix[lc] : max_suffix[rc];

      // 递推max_sub      
      max_sub[o] = better(max_sub[lc], max_sub[rc]); // 完全在左子树或者右子树
      max_sub[o] = better(max_sub[o], make_pair(max_suffix[lc], max_prefix[rc])); // 跨越中线
    }
  }

  Interval query_prefix(int o, int L, int R) {
    if(max_prefix[o] <= qR) return make_pair(L, max_prefix[o]);
    int M = L + (R-L)/2;
    int lc = o*2, rc = o*2+1;
    if(qR <= M) return query_prefix(lc, L, M);
    Interval i = query_prefix(rc, M+1, R);
    i.first = L;
    return better(i, make_pair(L, max_prefix[lc]));
  }

  Interval query_suffix(int o, int L, int R) {
    if(max_suffix[o] >= qL) return make_pair(max_suffix[o], R);
    int M = L + (R-L)/2;
    int lc = o*2, rc = o*2+1;
    if(qL > M) return query_suffix(rc, M+1, R);
    Interval i = query_suffix(lc, L, M);
    i.second = R;
    return better(i, make_pair(max_suffix[rc], R));
  }

  Interval query(int o, int L, int R) {
    if(qL <= L && R <= qR) return max_sub[o];
    int M = L + (R-L)/2;
    int lc = o*2, rc = o*2+1;
    if(qR <= M) return query(lc, L, M);
    if(qL > M) return query(rc, M+1, R);
    Interval i1 = query_prefix(rc, M+1, R); // 右半的前缀
    Interval i2 = query_suffix(lc, L, M); // 左半的后缀
    Interval i3 = better(query(lc, L, M), query(rc, M+1, R));
    return better(make_pair(i2.first, i1.second), i3);
  }
};

IntervalTree tree;

int main() {
  int kase = 0, n, a, Q;
  while(scanf("%d%d", &n, &Q) == 2) {
    prefix_sum[0] = 0;
    for(int i = 0; i < n; i++) {
      scanf("%d", &a);
      prefix_sum[i+1] = prefix_sum[i] + a;
    }
    tree.build(1, 1, n);
    printf("Case %d:\n", ++kase);
    while(Q--) {
      int L, R;
      scanf("%d%d", &L, &R);
      qL = L; qR = R;
      Interval ans = tree.query(1, 1, n);
      printf("%d %d\n", ans.first, ans.second);
    }
  }
  return 0;
}
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