计蒜客 41399 - Light bulbs

题目链接:ICPC 2019 上海网络赛B题

对区间 \([l,r]\) 离散化成四个点 \(l - 1\)\(l\)\(r-1\)\(r\),然后分块维护,渐进时间复杂度 \(O(T \times (4m \log 4m + 4m \sqrt{4m}))\)

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#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAX_N = 1000 + 5;
int T, n, m, d[MAX_N * 4], opL[MAX_N], opR[MAX_N], tot, dL[MAX_N], dR[MAX_N];
int bsz, bt, lb[MAX_N], rb[MAX_N], blk[MAX_N * 4], tag[MAX_N], a[MAX_N * 4];
int main() {
scanf("%d", &T);
for (int cs = 1; cs <= T; ++cs) {
scanf("%d%d", &n, &m);
tot = 0;
d[++tot] = 0; d[++tot] = n;
for (int i = 1; i <= m; ++i) {
scanf("%d%d", opL + i, opR + i);
d[++tot] = opL[i]; d[++tot] = opL[i] - 1;
d[++tot] = opR[i]; d[++tot] = opR[i] - 1;
}
sort(d + 1, d + tot + 1);
tot = unique(d + 1, d + tot + 1) - d - 1;
for (int i = 1; i <= m; ++i) {
dL[i] = lower_bound(d + 1, d + tot + 1, opL[i]) - d;
dR[i] = lower_bound(d + 1, d + tot + 1, opR[i]) - d;
}
bsz = ceil(int(sqrt(tot))); bt = 0;
int bl, br;
for (bl = 1; bl <= tot; bl = br + 1) {
br = min(tot, bl + bsz - 1); ++bt;
lb[bt] = bl; rb[bt] = br; tag[bt] = 0;
for (int i = bl; i <= br; ++i) {
blk[i] = bt; a[i] = 0;
}
}
int sb, eb;
for (int q = 1; q <= m; ++q) {
sb = blk[dL[q]]; eb = blk[dR[q]];
if (sb == eb) {
for (int i = dL[q]; i <= dR[q]; ++i) ++a[i];
continue;
}
for (int i = dL[q]; i <= rb[sb]; ++i) ++a[i];
for (int i = sb + 1; i < eb; ++i) ++tag[i];
for (int i = lb[eb]; i <= dR[q]; ++i) ++a[i];
}
int cnt = 0, tl, tr, len;
for (int i = 1; i <= tot; ++i) {
// printf("d[%d] = %d, a = %d\n", i, d[i], a[i] + tag[blk[i]]);
if ((a[i] + tag[blk[i]]) & 1) {
tl = d[i - 1] + 1;
tr = d[i];
len = max(tr - tl + 1, 0);
cnt += len;
}
}
printf("Case #%d: %d\n", cs, cnt);
}
return 0;
}