[bzoj1148][CTSC2007]挂缀pendant

大意:给定若干个任务,每个任务只能在给定时刻c[i]前开始,时长w[i],求最多能完成的任务.
可以看成:
有若干区间,给出长度len[i]和左端点l[i]的限制,选出尽量多使其互不重叠.
右端点限制r[i]=l[i]+len[i]
对于i < j,若r[i] < r[j]:如果j区间可以放置于i前,那么j区间一定也可以放置于i之后;
但是如果j区间不能放置于i前,j区间可能还是可以放置于i之后.
于是令所有的i < j都有r[i]

#include <cstdio>#include <queue>#include <algorithm>using namespace std;#define ll long long#define N 1000005struct node{    ll x,y;}a[N];bool cmp(node x,node y) {return x.x<y.x;}inline ll read(){    char ch=getchar();    ll x=0LL;    while ('0'>ch || ch>'9') ch=getchar();    while ('0'<=ch && ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();    return x;}priority_queue <ll> heap;int n,tot;int main(){    n=read();    for (int i=1;i<=n;++i) a[i]={read(),read()},a[i].x+=a[i].y;    sort(a+1,a+1+n,cmp);    ll s=0LL;    for (int i=1;i<=n;++i)    {        if (s+a[i].y<=a[i].x) {s+=a[i].y;++tot;heap.push(a[i].y);continue;}        if (heap.top()>a[i].y) s+=(a[i].y-heap.top()),heap.pop(),heap.push(a[i].y);    }    printf("%d\n%lld\n",tot,s);}