poj 3616 Milking Time

给出m个区间，每个区间有一个权值

选出若干个区间，使得区间两两不重合并且区间的距离大于r，区间的权值和最大

求权值和

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注意到区间只有1000个，所以离散化一下就可以O(m2)的做了

代码也很短

#include<vector>#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<stack>using namespace std;#define LL long longconst int maxn = 1123;int msp[maxn*2];int dp[maxn*2];struct Seg{    int l,r,v;};Seg seg[maxn];bool cmp(Seg a,Seg b){    return a.r < b.r;}int main(){    int n,m,r;    while(~scanf("%d %d %d",&n,&m,&r)){        int mlen = 0;        for(int i=0;i<m;i++){            scanf("%d %d %d",&seg[i].l,&seg[i].r,&seg[i].v);            msp[mlen++] = seg[i].l,msp[mlen++] = seg[i].r;        }        sort(msp,msp+mlen);        mlen = unique(msp,msp+mlen) - msp;        sort(seg,seg+m,cmp);        memset(dp,0,sizeof(dp));        for(int i=0;i<m;i++){            int lo = lower_bound(msp,msp+mlen,seg[i].r)-msp;            int bef = 0;            for(int j=0;msp[j]+r<=seg[i].l && j < mlen;j++){                bef = max(bef,dp[j]);            }            dp[lo] = max(dp[lo],bef + seg[i].v);        }        printf("%d\n",*max_element(dp,dp+mlen));    }    return 0;}