Codeforces Round #547 (Div. 3) F 贪心 + 离散化
https://codeforces.com/contest/1141/problem/F2
题意
一个大小为n的数组a[],问最多有多少个不相交的区间和相等
题解
- 离散化用值来做,贪心选择较前的区间
代码
#include<bits/stdc++.h>
#define M 5000005
#define ll long long
#define pb push_back
using namespace std;
struct N{int l,r;N(int l=0,int r=0):l(l),r(r){}};
int n,i,j,p,ans,cnt,u,r,sz=0;
ll a[2005],b[M];
vector<N>g[M];
int main(){
cin>>n;
for(i=1;i<=n;i++){
scanf("%lld",&a[i]);a[i]+=a[i-1];
for(j=i;j>=1;j--){
b[sz++]=a[i]-a[j-1];
}
}
sort(b,b+sz);sz=unique(b,b+sz)-b;
for(i=1;i<=n;i++){
for(j=1;j<=i;j++){
p=lower_bound(b,b+sz,a[i]-a[j-1])-b;
g[p].pb({j,i});
}
}
ans=1;
for(i=0;i<sz;i++){
r=g[i][0].r;cnt=1;
for(j=1;j<g[i].size();j++){
if(g[i][j].l>r){
cnt++;r=g[i][j].r;
}
}
if(cnt>=ans){
ans=cnt;u=i;
}
}
printf("%d\n%d %d\n",ans,g[u][0].l,g[u][0].r);
r=g[u][0].r;
for(i=1;i<g[u].size();i++){
if(g[u][i].l>r){
printf("%d %d\n",g[u][i].l,g[u][i].r);
r=g[u][i].r;
}
}
}
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