BZOJ 2752: [HAOI2012]高速公路(road) (线段树) - Gemini Boy - ACM之路~
BZOJ 2752: [HAOI2012]高速公路(road) (线段树)
链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2752
概率背景的题，不过稍微想想，稍微推一推，就能得到公式。然后注意到只要用线段树维护3个sum就行了，sum1 费用和， sum2 下标乘以费用的和 sum3下标的平方乘以费用的和
PS：。。。在写线段树的时候被一个傻逼问题bug住了好久，，调了好久。好吧 我就是煞笔
2版线段树的代码。
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100100;
long long sumo[maxn], sump[maxn];
struct SegNode {
int left, right;
long long sum1, sum2, sum3;
long long add;
int mid() {
return (left + right) >> 1;
}
long long size() {
return (right - left + 1);
}
};
struct SegmentTree {
SegNode tree[maxn*5];
void pushUp(int idx) {
tree[idx].sum1 = tree[idx<<1].sum1 + tree[idx<<1|1].sum1;
tree[idx].sum2 = tree[idx<<1].sum2 + tree[idx<<1|1].sum2;
tree[idx].sum3 = tree[idx<<1].sum3 + tree[idx<<1|1].sum3;
}
void pushDown(int idx) {
if (tree[idx].add) {
long long add = tree[idx].add;
int right1 = tree[idx<<1].right;
int left1 = tree[idx<<1].left;
int right2 = tree[idx<<1|1].right;
int left2 = tree[idx<<1|1].left;
tree[idx].add = 0;
tree[idx<<1].sum1 += tree[idx<<1].size() * add;
tree[idx<<1|1].sum1 += tree[idx<<1|1].size() * add;
tree[idx<<1].sum2 += (sumo[right1] - sumo[left1-1]) * add;
tree[idx<<1|1].sum2 += (sumo[right2] - sumo[left2-1]) * add;
tree[idx<<1].sum3 += (sump[right1] - sump[left1-1]) * add;
tree[idx<<1|1].sum3 += (sump[right2] - sump[left2-1]) * add;
tree[idx<<1].add += add;
tree[idx<<1|1].add += add;
}
}
void build(int left, int right, int idx) {
tree[idx].left = left;
tree[idx].right = right;
tree[idx].sum1 = 0;
tree[idx].sum2 = 0;
tree[idx].sum3 = 0;
tree[idx].add = 0;
if (left == right)
return ;
int mid = tree[idx].mid();
build(left, mid, idx<<1);
build(mid+1, right, idx<<1|1);
}
void update(int left, int right, int idx, long long value) {
if (tree[idx].left == left && tree[idx].right == right) {
tree[idx].add += value;
tree[idx].sum1 += tree[idx].size() * value;
tree[idx].sum2 += (sumo[right] - sumo[left-1]) * value;
tree[idx].sum3 += (sump[right] - sump[left-1]) * value;
return ;
}
pushDown(idx);
int mid = tree[idx].mid();
if (right <= mid)
update(left, right, idx<<1, value);
else if (left > mid)
update(left, right, idx<<1|1, value);
else {
update(left, mid, idx<<1, value);
update(mid+1, right, idx<<1|1, value);
}
pushUp(idx);
}
long long query(int left, int right, int idx, int L, int R) {
if (tree[idx].left == left && tree[idx].right == right) {
//printf("%d %d %d\n", idx, left, right);
return (R + L - 1) * tree[idx].sum2 - tree[idx].sum3 -
((long long)L * R - R) * tree[idx].sum1;
}
pushDown(idx);
int mid = tree[idx].mid();
if (right <= mid)
return query(left, right, idx<<1, L, R);
else if (left > mid)
return query(left, right, idx<<1|1, L, R);
else {
return (query(left, mid, idx<<1, L, R) + query(mid+1, right, idx<<1|1, L, R));
}
}
};
SegmentTree tree;
int N, M;
long long gcd(long long a, long long b) {
return b ? gcd(b, a % b) : a;
}
void init() {
memset(sumo, 0, sizeof(sumo));
memset(sump, 0, sizeof(sump));
for (int i = 1; i < maxn; i++)
sumo[i] = sumo[i-1] + i;
for (int i = 1; i < maxn; i++)
sump[i] = sump[i-1] + (long long)i * i;
scanf("%d %d", &N, &M);
tree.build(1, N, 1);
}
void work() {
char op[19];
for (int i = 0; i < M; i++) {
scanf("%s", op);
if (op[0] == 'C') {
int x, y;
long long z;
scanf("%d %d %lld", &x, &y, &z);
tree.update(x, y-1, 1, z);
} else {
int x, y;
scanf("%d %d", &x, &y);
long long tmp = tree.query(x, y, 1, x, y);
long long tem = (long long)(y - x + 1) * (y - x) / 2;
long long ans = gcd(tmp, tem);
printf("%lld/%lld\n", tmp / ans, tem / ans);
}
}
}
int main() {
init();
work();
return 0;
}
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 100100;
long long sumo[maxn], sump[maxn];
struct SegNode {
int left, right;
long long sum1, sum2, sum3;
long long add;
};
struct SegmentTree {
SegNode tree[maxn*5];
void pushUp(int idx) {
tree[idx].sum1 = tree[idx<<1].sum1 + tree[idx<<1|1].sum1;
tree[idx].sum2 = tree[idx<<1].sum2 + tree[idx<<1|1].sum2;
tree[idx].sum3 = tree[idx<<1].sum3 + tree[idx<<1|1].sum3;
}
void pushDown(int idx, int left, int right) {
if (tree[idx].add) {
long long add = tree[idx].add;
tree[idx].add = 0;
tree[idx].sum1 += (right - left + 1) * add;
tree[idx].sum2 += (sumo[right] - sumo[left-1]) * add;
tree[idx].sum3 += (sump[right] - sump[left-1]) * add;
if (left == right) return ;
tree[idx<<1].add += add;
tree[idx<<1|1].add += add;
}
}
void update(int L, int R, int left, int right, int idx, long long value) {
pushDown(idx, L, R);
if (L >= left && R <= right) {
tree[idx].add += value;
return ;
}
int mid = (L + R) >> 1;
if (left <= mid)
update(L, mid, left, right, idx<<1, value);
if (right >= mid+1)
update(mid+1, R, left, right, idx<<1|1, value);
pushDown(idx<<1, L, mid);
pushDown(idx<<1|1, mid+1, R);
pushUp(idx);
}
long long query(int L, int R, int left, int right, int idx) {
pushDown(idx, L, R);
if (L >= left && R <= right) {
printf("%d %d %d\n", idx, left, right);
return (right + left - 1) * tree[idx].sum2 - tree[idx].sum3 -
((long long)left * right - right) * tree[idx].sum1;
}
int mid = (L + R) >> 1;
long long ans = 0;
if (left <= mid)
ans += query(L, mid, left, right, idx<<1);
if (right >= mid + 1)
ans += query(mid+1, R, left, right, idx<<1|1);
return ans;
}
};
SegmentTree tree;
int N, M;
long long gcd(long long a, long long b) {
return b ? gcd(b, a % b) : a;
}
void init() {
memset(sumo, 0, sizeof(sumo));
memset(sump, 0, sizeof(sump));
for (int i = 1; i < maxn; i++)
sumo[i] = sumo[i-1] + i;
for (int i = 1; i < maxn; i++)
sump[i] = sump[i-1] + (long long)i * i;
scanf("%d %d", &N, &M);
}
void work() {
char op[19];
for (int i = 0; i < M; i++) {
scanf("%s", op);
if (op[0] == 'C') {
int x, y;
long long z;
scanf("%d %d %lld", &x, &y, &z);
tree.update(1, N, x, y-1, 1, z);
} else {
int x, y;
scanf("%d %d", &x, &y);
long long tmp = tree.query(1, N, x, y, 1);
long long tem = (long long)(y - x + 1) * (y - x) / 2;
long long ans = gcd(tmp, tem);
printf("%lld/%lld\n\n", tmp / ans, tem / ans);
}
}
}
int main() {
init();
work();
return 0;
}
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