Best Financing（HD4833）

每笔收入产生的收益是独立的。

计算所有点的收益率，累计。

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

struct dd{
    int d, e;
}ndd[2505];
struct ddd{
    int s, f, r;
}mddd[2505];
bool dddcomp(ddd a, ddd b)
{
    return a.s < b.s;
}
int rate[100005];

int max(int a, int b)
{
    return a>b ? a : b;
}
int main()
{
    int t, n, m;
    scanf("%d", &t);
    for (int w = 1; w <= t; ++w){
        memset(rate, 0, sizeof(rate));
        scanf("%d %d", &n, &m);
        for (int i = 0; i < n; ++i)
            scanf("%d %d", &ndd[i].d, &ndd[i].e);
        for (int i = 0; i < m; ++i)
            scanf("%d %d %d", &mddd[i].s, &mddd[i].f, &mddd[i].r);
        sort(mddd, mddd + m, dddcomp);        //读取数据并排序
        //求rate[],rate[i]=max(rate[i+1],rate[j]+r[i-j]),只需考虑 r[i-j]存在的情况
        int pos = m - 1;
        for (int i = mddd[m - 1].s; i >= mddd[0].s; --i){
            int mx = rate[i + 1];
            while (pos >= 0 && i == mddd[pos].s){
                mx = max(mx, mddd[pos].r + rate[mddd[pos].f]);
                --pos;
            }
            rate[i] = mx;
        }
        for (int i = mddd[0].s - 1; i >= 0; --i)
            rate[i] = rate[i + 1];
        //分别计算收入的收益
        long long sum = 0;
        for (int i = 0; i < n; ++i){
            sum += rate[ndd[i].d] * ndd[i].e;
        }
        printf("Case #%d:
%.2lf
", w, (double)sum / 100.0);
    }
}