Codeforces 1313D Happy New Year
题目链接
https://codeforces.com/contest/1313/problem/D
题目大意
一共 m 个点，用 n 条线段去覆盖，问被覆盖奇数次的点最多有多少。保证一个点最多被 8 条不同的线段覆盖。
题目分析
注意到每个点最多被 8 条线段覆盖，猜测复杂度可能和 2 ^ 8 有关。考虑到如果一条线段同时覆盖 i 和 i + 1 两个点，那么这条线段选不选在 i 时就可以确定。因此考虑使用 DP 解决问题。
设 F[i][S] 为已经考虑到第 i 个点，选择线段集合 S。显然已经结束的线段和还没有出现的线段不需要考虑，因此 S 可以只记录覆盖第 i 个点的几个线段。显然 i – 1 已经选择的线段必选。所以可以枚举重叠部分，在考虑不重叠部分进行转移，很容易就可以解决。由于 m 很大，因此还需要对区间进行离散化，将覆盖线段相同的点分在一块儿。复杂度即可达到题目要求。
代码
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int n, m, k;
pair <int, int> a[maxn];
vector <pair <int, int> > vec;
vector <int> c[maxn];
int f[maxn][(1<<9)];
int g[(1<<9)];
int v[10];
int u[10];
int main(){
int i, j;
int x, y;
int l, r;
scanf("%d%d%d", &n, &m, &k);
for(i=1;i<=n;i++){
scanf("%d%d", &l, &r);
a[i] = make_pair(l, r);
}
sort(a + 1, a + n + 1);
x = 1;
priority_queue <int> q;
a[n + 1] = make_pair(m + 1, m + 1);
for(i=1;i<=n+1;i++){
while(q.size()){
if(-q.top() < a[i].first){
if(-q.top() >= x){
vec.push_back(make_pair(x, -q.top()));
x = -q.top() + 1;
}
q.pop();
}else{
break;
}
}
if(x < a[i].first){
vec.push_back(make_pair(x, a[i].first - 1));
x = a[i].first;
}
q.push(-a[i].second);
}
j = 0;
for(i=1;i<=n;i++){
while(j < vec.size() and a[i].first > vec[j].first) j++;
for(x=j;x<vec.size() and a[i].second>=vec[x].second;x++){
c[x].push_back(i);
}
}
int ans = 0;
for(j=0;j<(1<<c[0].size());j++){
f[0][j] = (__builtin_popcount(j) & 1) * (vec[0].second - vec[0].first + 1);
ans = max(ans, f[0][j]);
}
for(i=1;i<vec.size();i++){
memset(v, -1, sizeof(v));
memset(g, 0, sizeof(g));
int z = 0;
for(x=0;x<c[i-1].size();x++){
for(y=0;y<c[i].size();y++){
if(c[i - 1][x] == c[i][y]) v[x] = y, z |= (1 << y);
}
}
for(j=0;j<(1<<c[i-1].size());j++){
int tmp = 0;
for(x=0;x<c[i-1].size();x++){
if(((j >> x) & 1) and v[x] != -1){
tmp |= (1 << v[x]);
}
}
g[tmp] = max(g[tmp], f[i - 1][j]);
}
for(j=0;j<(1<<c[i].size());j++){
int tmp = (j & z);
f[i][j] = max(f[i][j], g[tmp] + (__builtin_popcount(j) & 1) * (vec[i].second - vec[i].first + 1));
ans = max(ans, f[i][j]);
}
}
printf("%d\n", ans);
return 0;
}
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int n, m, k;
pair <int, int> a[maxn];
vector <pair <int, int> > vec;
vector <int> c[maxn];
int f[maxn][(1<<9)];
int g[(1<<9)];
int v[10];
int u[10];
int main(){
int i, j;
int x, y;
int l, r;
scanf("%d%d%d", &n, &m, &k);
for(i=1;i<=n;i++){
scanf("%d%d", &l, &r);
a[i] = make_pair(l, r);
}
sort(a + 1, a + n + 1);
x = 1;
priority_queue <int> q;
a[n + 1] = make_pair(m + 1, m + 1);
for(i=1;i<=n+1;i++){
while(q.size()){
if(-q.top() < a[i].first){
if(-q.top() >= x){
vec.push_back(make_pair(x, -q.top()));
x = -q.top() + 1;
}
q.pop();
}else{
break;
}
}
if(x < a[i].first){
vec.push_back(make_pair(x, a[i].first - 1));
x = a[i].first;
}
q.push(-a[i].second);
}
j = 0;
for(i=1;i<=n;i++){
while(j < vec.size() and a[i].first > vec[j].first) j++;
for(x=j;x<vec.size() and a[i].second>=vec[x].second;x++){
c[x].push_back(i);
}
}
int ans = 0;
for(j=0;j<(1<<c[0].size());j++){
f[0][j] = (__builtin_popcount(j) & 1) * (vec[0].second - vec[0].first + 1);
ans = max(ans, f[0][j]);
}
for(i=1;i<vec.size();i++){
memset(v, -1, sizeof(v));
memset(g, 0, sizeof(g));
int z = 0;
for(x=0;x<c[i-1].size();x++){
for(y=0;y<c[i].size();y++){
if(c[i - 1][x] == c[i][y]) v[x] = y, z |= (1 << y);
}
}
for(j=0;j<(1<<c[i-1].size());j++){
int tmp = 0;
for(x=0;x<c[i-1].size();x++){
if(((j >> x) & 1) and v[x] != -1){
tmp |= (1 << v[x]);
}
}
g[tmp] = max(g[tmp], f[i - 1][j]);
}
for(j=0;j<(1<<c[i].size());j++){
int tmp = (j & z);
f[i][j] = max(f[i][j], g[tmp] + (__builtin_popcount(j) & 1) * (vec[i].second - vec[i].first + 1));
ans = max(ans, f[i][j]);
}
}
printf("%d\n", ans);
return 0;
}
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int n, m, k;
pair <int, int> a[maxn];
vector <pair <int, int> > vec;
vector <int> c[maxn];
int f[maxn][(1<<9)];
int g[(1<<9)];
int v[10];
int u[10];
int main(){
int i, j;
int x, y;
int l, r;
scanf("%d%d%d", &n, &m, &k);
for(i=1;i<=n;i++){
scanf("%d%d", &l, &r);
a[i] = make_pair(l, r);
}
sort(a + 1, a + n + 1);
x = 1;
priority_queue <int> q;
a[n + 1] = make_pair(m + 1, m + 1);
for(i=1;i<=n+1;i++){
while(q.size()){
if(-q.top() < a[i].first){
if(-q.top() >= x){
vec.push_back(make_pair(x, -q.top()));
x = -q.top() + 1;
}
q.pop();
}else{
break;
}
}
if(x < a[i].first){
vec.push_back(make_pair(x, a[i].first - 1));
x = a[i].first;
}
q.push(-a[i].second);
}
j = 0;
for(i=1;i<=n;i++){
while(j < vec.size() and a[i].first > vec[j].first) j++;
for(x=j;x<vec.size() and a[i].second>=vec[x].second;x++){
c[x].push_back(i);
}
}
int ans = 0;
for(j=0;j<(1<<c[0].size());j++){
f[0][j] = (__builtin_popcount(j) & 1) * (vec[0].second - vec[0].first + 1);
ans = max(ans, f[0][j]);
}
for(i=1;i<vec.size();i++){
memset(v, -1, sizeof(v));
memset(g, 0, sizeof(g));
int z = 0;
for(x=0;x<c[i-1].size();x++){
for(y=0;y<c[i].size();y++){
if(c[i - 1][x] == c[i][y]) v[x] = y, z |= (1 << y);
}
}
for(j=0;j<(1<<c[i-1].size());j++){
int tmp = 0;
for(x=0;x<c[i-1].size();x++){
if(((j >> x) & 1) and v[x] != -1){
tmp |= (1 << v[x]);
}
}
g[tmp] = max(g[tmp], f[i - 1][j]);
}
for(j=0;j<(1<<c[i].size());j++){
int tmp = (j & z);
f[i][j] = max(f[i][j], g[tmp] + (__builtin_popcount(j) & 1) * (vec[i].second - vec[i].first + 1));
ans = max(ans, f[i][j]);
}
}
printf("%d\n", ans);
return 0;
}