51nod 1611 金牌赛事(动态规划 + 线段树)

分析：看起来有点像最大权闭合图，然而复杂度太高。。。

正解是dp，设dp[i]为考虑前i条路的最大收益，则dp[i]=max{dp[j] - cost[j+1][i] + earn[j+1][i]}，0<=j<=i-1，earn[j+1][i]表示在[j+1，i]之间的比赛，是个O(n^2)的dp.

接下来考虑优化这个dp.维护一个线段树，当扫到i时，树中满足a[j] = dp[j] - cost[j+1][i] + earn[j+1][i]，先将比赛按右端点排序，然后每扫到一个右端点为i的比赛，将所有j<l的a[j] 加上earn，将所有j<i的a[i]减去c[i]，然后更新dp[i]为[0,i-1]的最大值。复杂度为O(nlogn)。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
const ll inf = 1e18;
struct Contest{
    int l,r;
    ll earn;
}p[maxn];
int n, m;
//bool Cmp(Pair a,Pair b){return a.r-a.l<b.r-b.l;}
bool Cmp(Contest a, Contest b){return a.r < b.r;}
class segTree{
    ll a[maxn],s[maxn*4],tag[maxn*4];
public:
    segTree(){memset(a,0,sizeof(a));}
    void build(int node,int begin,int end){
        tag[node]=0;
        if(begin==end){
            s[node]=a[begin];
        }else{
            build(2*node,begin,(begin+end)/2);
            build(2*node+1,(begin+end)/2+1,end);
            s[node]=max(s[2*node],s[2*node+1]);
        }
    }
    void pushdown(int node,int begin,int end){
        if(begin==end){
            s[node]+=tag[node];
            a[begin]=s[node];
        }else{
            tag[2*node]+=tag[node];
            tag[2*node+1]+=tag[node];
            s[node]+=tag[node];
        }
        tag[node]=0;
    }
    void Change(int node,int begin,int end,int left,int right,ll add){
        if(begin>=left&&end<=right){
            tag[node]+=add;
        }else{
            pushdown(node,begin,end);
            int m=(begin+end)/2;
            if(!(right<begin||left>m)){
                Change(2*node,begin,m,left,right,add);
            }
            if(!(right<m+1||left>end)){
                Change(2*node+1,m+1,end,left,right,add);
            }
            s[node] = max(s[2 * node] + tag[2 * node], s[2 * node + 1] + tag[2 * node + 1]);
        }
    }
    ll query(int node,int begin,int end,int left,int right){
        pushdown(node,begin,end);
        if(begin > right || end < left)return -inf;
        if(begin==end)return a[begin];
        if(begin >= left && end <= right)return s[node];
        ll q1 = query(2 * node, begin, (begin + end) / 2, left, right);
        ll q2 = query(2 * node + 1, (begin + end) / 2 + 1, end, left, right);
        return max(q1, q2);
    }
    void Print(){
        for(int i = 0; i <= n; i++){
            cout<<query(1, 0, n, i, i)<<' ';
        }
        cout<<endl;
    }
}st;

ll c[maxn], f[maxn];
int main(){
//    freopen("e:\in.txt","r",stdin);
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n; i++)scanf("%lld",&c[i]);
    for(int i = 0; i < m; i++)scanf("%d%d%lld",&p[i].l,&p[i].r,&p[i].earn);
    sort(p, p + m, Cmp);
    int idx = 0;
    st.build(1, 0, n);
    f[0] = 0;
    for(int i = 1; i <= n; i++){
        while(idx < m && p[idx].r <= i){
            st.Change(1, 0, n, 0, p[idx].l - 1, p[idx].earn);
//            st.Print();
            idx++;
        }
        st.Change(1, 0, n, 0, i - 1, -c[i]);
//        st.Print();
        f[i] = st.query(1, 0, n, 0, i - 1);
        f[i] = max(f[i - 1], f[i]);
        st.Change(1, 0, n, i, i, f[i]);
//        st.Print();
//        cout<<"*********"<<endl;
    }
//    for(int i = 0; i <= n; i++)cout<<f[i]<<endl;
    cout<<f[n]<<endl;
    return 0;
}