http://acm.hdu.edu.cn/showproblem.php?pid=6012
我们希望能够快速算出，对于每一个温度，都能够算出它在这n颗植物中，能得到多少价值。
那么，对于第i科植物，在[0, L[i] - 1]这些温度中，得到的价值是低温那个价值，同理在[L[i], R[i]]中，和[R[i], mx]中，
那么可以用O(1)打标记的思路去完成。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
map<int, LL>book;
void work() {
book.clear();
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
int L, R, a, b, c;
scanf("%d%d%d%d%d", &L, &R, &a, &b, &c);
book[0] += c;
book[L << 1] += a - c;
book[(R << 1) + 1] += b - a;
}
LL ans = 0;
LL tans = 0;
for (map<int, LL> :: iterator it = book.begin(); it != book.end(); ++it) {
tans += it->second;
ans = max(ans, tans);
}
printf("%I64d\n", ans);
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
int t;
scanf("%d", &t);
while (t--) work();
return 0;
}
View Code
同样可以用线段树完成，就是先把坐标离散了，然后区间更新和上面一样的东西。
线段树维护最大值，seg[cur]表示这课树覆盖的区间，其中某个温度能取得的最大值。
但是我一直wa，不知为何。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
#define lson L, mid, cur << 1
#define rson mid + 1, R, cur << 1 | 1
#define root 1, mx, 1
const int maxn = 500000 + 20;
int L[maxn], R[maxn];
LL a[maxn], b[maxn], c[maxn];
LL add[maxn << 3], seg[maxn << 3];
vector<int>da;
void pushDown(int cur) {
if (add[cur]) {
add[cur << 1] += add[cur];
add[cur << 1 | 1] += add[cur];
seg[cur << 1 | 1] += add[cur];
seg[cur << 1] += add[cur];
add[cur] = 0;
}
}
void pushUp(int cur) {
seg[cur] = max(seg[cur << 1], seg[cur << 1 | 1]);
assert(seg[cur] >= 0);
}
void build(int L, int R, int cur) {
seg[cur] = add[cur] = 0;
if (L == R) {
return;
}
int mid = (L + R) >> 1;
build(lson);
build(rson);
pushUp(cur);
}
void upDate(int be, int en, LL val, int L, int R, int cur) {
if (L >= be && R <= en) {
seg[cur] += val;
add[cur] += val;
return;
}
pushDown(cur);
int mid = (L + R) >> 1;
if (be <= mid) upDate(be, en, val, lson);
if (en > mid) upDate(be, en, val, rson);
pushUp(cur);
}
LL query(int be, int en, int L, int R, int cur) {
if (L >= be && R <= en) return seg[cur];
pushDown(cur);
LL ans = 0;
int mid = (L + R) >> 1;
if (be <= mid) ans += query(be, en, lson);
if (en > mid) ans += query(be, en, rson);
return ans;
}
void work() {
da.clear();
int n;
scanf("%d", &n);
da.push_back(-inf);
da.push_back(-inf);
da.push_back(-inf);
for (int i = 1; i <= n; ++i) {
scanf("%d%d%I64d%I64d%I64d", &L[i], &R[i], &a[i], &b[i], &c[i]);
assert(L[i] <= R[i]);
if (L[i] > R[i]) swap(L[i], R[i]);
L[i] *= 2;
R[i] *= 2;
da.push_back(L[i]);
da.push_back(R[i]);
}
sort(da.begin(), da.end());
int gg = da.size();
for (int i = 3; i < gg; ++i) {
if (da[i] - da[i - 1] == 2) {
da.push_back(da[i] - 1);
}
}
sort(da.begin(), da.end());
int mx = 0;
for (int i = 1; i <= n; ++i) {
L[i] = lower_bound(da.begin(), da.end(), L[i]) - da.begin();
R[i] = lower_bound(da.begin(), da.end(), R[i]) - da.begin();
// cout << L[i] << " " << R[i] << endl;
mx = max(mx, R[i] + 2);
}
// cout << endl;
build(root);
for (int i = 1; i <= n; ++i) {
upDate(1, L[i] - 1, c[i], root);
upDate(L[i], R[i], a[i], root);
upDate(R[i] + 1, mx, b[i], root);
}
cout << query(1, mx, root) << endl;
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
int t;
scanf("%d", &t);
while (t--) work();
return 0;
}
View Code
原来是我离散化错了
一开始的时候，认为[70, 73] [74, 76]这样，73和74相邻，那么乘以2后，变成146 148，那么相隔2的时候才添加些元素进去，比如添加147进去隔着，这样离散化。
这样有bug（还没想到有什么bug）
现在的思路是：乘以2后，把L[i] + 1和R[i] + 1也放进去，这样就不会有挨着了
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
#define lson L, mid, cur << 1
#define rson mid + 1, R, cur << 1 | 1
#define root 1, mx, 1
const int maxn = 500000 + 20;
int L[maxn], R[maxn];
int a[maxn], b[maxn], c[maxn];
LL add[maxn << 3], seg[maxn << 3];
vector<int>da;
void pushDown(int cur) {
if (add[cur]) {
add[cur << 1] += add[cur];
add[cur << 1 | 1] += add[cur];
seg[cur << 1 | 1] += add[cur];
seg[cur << 1] += add[cur];
add[cur] = 0;
}
}
void pushUp(int cur) {
seg[cur] = max(seg[cur << 1], seg[cur << 1 | 1]);
}
void build(int L, int R, int cur) {
seg[cur] = add[cur] = 0;
if (L == R) {
return;
}
int mid = (L + R) >> 1;
build(lson);
build(rson);
pushUp(cur);
}
void upDate(int be, int en, LL val, int L, int R, int cur) {
if (L >= be && R <= en) {
seg[cur] += val;
add[cur] += val;
return;
}
pushDown(cur);
int mid = (L + R) >> 1;
if (be <= mid) upDate(be, en, val, lson);
if (en > mid) upDate(be, en, val, rson);
pushUp(cur);
}
void work() {
da.clear();
int n;
scanf("%d", &n);
da.push_back(-inf);
da.push_back(-inf);
da.push_back(-inf);
for (int i = 1; i <= n; ++i) {
scanf("%d%d%d%d%d", &L[i], &R[i], &a[i], &b[i], &c[i]);
L[i] *= 2;
R[i] *= 2;
da.push_back(L[i]);
da.push_back(L[i] + 1);
da.push_back(R[i] + 1);
da.push_back(R[i]);
}
sort(da.begin(), da.end());
int mx = 0;
for (int i = 1; i <= n; ++i) {
L[i] = lower_bound(da.begin(), da.end(), L[i]) - da.begin();
R[i] = lower_bound(da.begin(), da.end(), R[i]) - da.begin();
// cout << L[i] << " " << R[i] << endl;
mx = max(mx, R[i] + 2);
}
// cout << endl;
build(root);
for (int i = 1; i <= n; ++i) {
upDate(1, L[i] - 1, c[i], root);
upDate(L[i], R[i], a[i], root);
upDate(R[i] + 1, mx, b[i], root);
}
printf("%I64d\n", seg[1]);
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
int t;
scanf("%d", &t);
while (t--) work();
return 0;
}
View Code
现在的方法是直接
