题目链接
题解
看这(O(n^3))的数据范围，可以想到区间(dp)
发现同一个会场的活动可以重叠，所以暴力求出(num[l][r])表示离散化后([l,r])的完整活动数
我们的目标求出(F[l][r])表示([l,r])必须选时，二者的最小值
我们不妨令(A)选了([l,r])，我们枚举(A)在([1,l - 1])和([r + 1,tot])各选了多少，求出此时(B)能选的最大值
如果我们能求出(f[i][j])表示([1,i])中(A)选了(j)个时(B)能选的最大值，(g[i][j])同理表示后缀，就可以求出(F[l][r])
(f[i][j])可以枚举断点(k)从而(O(n^3))转移
[f[i][j] = max{f[k][j] + num[k + 1][i],f[i][j - num[k + 1][i]]}
]
(F[l][r])的转移是(O(n^4))，这个过程中我们(O(n^2))枚举了(A)在两端的选取个数
感性理解一下，当(A)在左端选多时，为了使答案更优，右端应该选少一些，所以左端增大的同时右端应该是单调变化的
用一个指针维护右端即可(O(n^3))
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u]; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 405,maxm = 100005,INF = 100000000;
const double eps = 1e-9;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
return flag ? out : -out;
}
int n,L[maxn],R[maxn],b[maxn],bi,tot;
int f[maxn][maxn],g[maxn][maxn],num[maxn][maxn];
int F[maxn][maxn];
inline int getn(int x){return lower_bound(b + 1,b + 1 + tot,x) - b;}
inline int cal(int l,int r,int x,int y){
return min(x + y + num[l][r],f[l - 1][x] + g[r + 1][y]);
}
void work(){
for (int j = 1; j <= n; j++) f[0][j] = -INF;
for (int i = 1; i <= tot; i++)
for (int j = 0; j <= n; j++){
f[i][j] = -INF;
for (int k = 0; k < i; k++){
int tmp = -INF;
if (num[1][k] >= j) tmp = max(tmp,f[k][j] + num[k + 1][i]);
if (num[k + 1][i] <= j) tmp = max(tmp,f[k][j - num[k + 1][i]]);
f[i][j] = max(f[i][j],tmp);
}
}
for (int j = 1; j <= n; j++) g[tot + 1][j] = -INF;
for (int i = tot; i; i--)
for (int j = 0; j <= n; j++){
g[i][j] = -INF;
for (int k = tot + 1; k > i; k--){
int tmp = -INF;
if (num[k][tot] >= j) tmp = max(tmp,g[k][j] + num[i][k - 1]);
if (num[i][k - 1] <= j) tmp = max(tmp,g[k][j - num[i][k - 1]]);
g[i][j] = max(g[i][j],tmp);
}
}
//REP(i,tot) REP(j,n) printf("f[%d][%d] = %d
",i,j,f[i][j]);
int ans = 0;
for (int i = 0; i <= n; i++)
ans = max(ans,min(i,f[tot][i]));
printf("%d
",ans);
for (int len = tot; len; len--)
for (int l = 1; l + len - 1 <= tot; l++){
int r = l + len - 1,y = num[r + 1][tot];
F[l][r] = max(F[l - 1][r],F[l][r + 1]);
for (int x = 0; x <= num[1][l - 1]; x++){
while (y > 0 && cal(l,r,x,y - 1) >= cal(l,r,x,y)) y--;
F[l][r] = max(F[l][r],cal(l,r,x,y));
}
}
REP(i,n) printf("%d
",F[L[i]][R[i]]);
}
int main(){
n = read();
REP(i,n){
b[++bi] = L[i] = read();
b[++bi] = R[i] = read() + L[i] - 1;
}
sort(b + 1,b + 1 + bi); tot = 1;
for (int i = 2; i <= bi; i++) if (b[i] != b[tot]) b[++tot] = b[i];
REP(i,n){
L[i] = getn(L[i]),R[i] = getn(R[i]);
for (int l = L[i]; l; l--)
for (int r = R[i]; r <= tot; r++)
num[l][r]++;
}
//REP(i,n) printf("[%d,%d]
",L[i],R[i]); puts("");
//printf("%d
",num[3][5]);
work();
return 0;
}