hdu 5454 Excited Database（线段树）

题目链接：hdu 5454 Excited Database

解题思路

维护两科线段树，一棵i+j，一棵i-j。因为对角线上的值一定都是相同的，只是查询时的个数不一样，但是根据区间范围可以计算出来每条对角线的个数。每个树维护单一节点权值和s，从右边递增的和r，从左边递增的和l。查询的时候将矩形分成3份。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 400005;
typedef long long ll;
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)

struct SegTree {
int lc[maxn<<2], rc[maxn<<2];
ll S[maxn<<2], L[maxn<<2], R[maxn<<2], T[maxn<<2];

int length(int u) {
return rc[u] - lc[u] + 1;
}

void maintain(int u, ll a) {
T[u] += a;
S[u] += a * length(u);
ll add = a * (length(u) + 1) * length(u) / 2;
L[u] += add;
R[u] += add;
}

void pushup(int u) {
S[u] = S[lson(u)] + S[rson(u)];
L[u] = L[lson(u)] + L[rson(u)] + S[rson(u)] * length(lson(u));
R[u] = R[rson(u)] + R[lson(u)] + S[lson(u)] * length(rson(u));
}

void pushdown(int u) {
if (T[u]) {
maintain(lson(u), T[u]);
maintain(rson(u), T[u]);
T[u] = 0;
}
}

void build (int u, int l, int r) {
lc[u] = l, rc[u] = r;
S[u] = L[u] = R[u] = T[u] = 0;

if (l == r) return;

int mid = (l+r)>>1;
build (lson(u), l, mid);
build (rson(u), mid+1, r);
pushup(u);
}

void modify(int u, int l, int r, int v) {
if (l <= lc[u] && rc[u] <= r) {
maintain(u, v);
return;
}

pushdown(u);
int mid = (lc[u] + rc[u]) >> 1;
if (l <= mid) modify(lson(u), l, r, v);
if (r > mid) modify(rson(u), l, r, v);
pushup(u);
}

ll query(int u, int l, int r) {
if (l <= lc[u] && rc[u] <= r)
return S[u];

pushdown(u);
int mid = (lc[u] + rc[u]) >> 1;
ll ret = 0;
if (l <= mid) ret += query(lson(u), l, r);
if (r > mid) ret += query(rson(u), l, r);
pushup(u);
return ret;
}

ll queryLeft(int u, int l, int r) {
if (l <= lc[u] && rc[u] <= r)
return L[u] + S[u] * (lc[u] - l);

pushdown(u);
int mid = (lc[u] + rc[u]) >> 1;
ll ret = 0;
if (l <= mid) ret += queryLeft(lson(u), l, r);
if (r > mid) ret += queryLeft(rson(u), l, r);
pushup(u);
return ret;
}

ll queryRight(int u, int l, int r) {
if (l <= lc[u] && rc[u] <= r)
return R[u] + S[u] * (r - rc[u]);

pushdown(u);
int mid = (lc[u] + rc[u]) >> 1;
ll ret = 0;
if (l <= mid) ret += queryRight(lson(u), l, r);
if (r > mid) ret += queryRight(rson(u), l, r);
pushup(u);
return ret;
}
}A, D;

int N, M;

int main () {
int cas;
scanf("%d", &cas);
for (int kcas = 1; kcas <= cas; kcas++) {
scanf("%d%d", &N, &M);
A.build(1, 2, 2 * N);
D.build(1, 1, 2 * N - 1);
//      A.build(1, 1, 2 * N);
//      D.build(1, 1, 2 * N);

printf("Case #%d:\n", kcas);

int t, l, r, x1, y1, x2, y2;
while (M--) {
scanf("%d", &t);
if (t == 1) {
scanf("%d%d", &l, &r);
A.modify(1, l, r, 1);
} else if (t == 2) {
scanf("%d%d", &l, &r);
D.modify(1, l + N, r + N, 1);
} else {
scanf("%d%d%d%d", &x1, &x2, &y1, &y2);
ll ans = 0, t = min(x2-x1, y2-y1) + 1;;

l = x2 + y1, r = x1 + y2;
if (l > r) swap(l, r);
ans += A.query(1, l, r) * t;
ans += A.queryLeft(1, x1+y1, l-1);
ans += A.queryRight(1, r+1, x2+y2);

l = x1-y1+N, r = x2-y2+N;
if (l > r) swap(l, r);
ans += D.query(1, l, r) * t;
ans += D.queryLeft(1, x1-y2+N, l-1);
ans += D.queryRight(1, r+1, x2-y1+N);
printf("%lld\n", ans);
}
}
}
return 0;
}