题目链接：
题目分析：
\(dp\)，思路和题解大部分有点区别，建议按洛谷题解写，这个有点丑
看成\(0/1\)背包之后发现是菜题
先假设牛把有命吃到的垃圾全都食下去了，然后记成生命值按时间消耗
设\(dp[i][j]\)表示处理到第\(i\)个垃圾，生命值还剩\(j\)时的最高高度
具体方程看代码，懒得打\(LaTeX\)了
代码：
#include<bits/stdc++.h>
#define N (100 + 10)
using namespace std;
inline int read() {
int cnt = 0, f = 1; char c = getchar();
while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();}
return cnt * f;
}
const int INF = (1 << 30);
int d, g, dp[N][N * 10], ans = (1 << 30), tmp = 0;
int vis[N][N * 10];
int lim;
struct node {
int t, f, h;
}a[N];
bool cmp(node a, node b) {
return a.t < b.t;
}
int main() {
// freopen("1.txt", "r", stdin);
d = read(), g = read();
for (register int i = 1; i <= g; ++i)
a[i].t = read(), a[i].f = read(), a[i].h = read();
sort(a + 1, a + g + 1, cmp);
lim = 10;
vis[0][10] = 1;
for (register int i = 1; i <= g; ++i) {
if (a[i].t > lim) break;
lim += a[i].f;
}
for (register int i = 1; i <= g; ++i) {
tmp += a[i].t - a[i - 1].t;
for (register int j = lim; j >= 0; --j) {
int delta = a[i].t - a[i - 1].t;
if (vis[i - 1][j + delta]) dp[i][j] = max(dp[i][j], dp[i - 1][j + delta] + a[i].h), vis[i][j] = 1;
if (vis[i - 1][j + delta]) dp[i][j + a[i].f] = max(dp[i][j + a[i].f], dp[i - 1][j + delta]), vis[i][j + a[i].f] = 1;
if (dp[i][j] >= d) {ans = a[i].t; break;}
}
if (ans != INF) break;
}
if (ans != INF) printf("%d", ans);
else printf("%d", lim);
return 0;
}