luogu P1156 垃圾陷阱

传送门

背包

每种物品有两种方法 也就是不叠高一定吃

所以dp[j]表示高度为j的时候能活多久 初值全是负的

dp[j]<0表示还不能叠到这个高度

所以当j+a[i].h>V 的时候 也就是i之前的东西能保证叠到j而且能活 说明逃出去了

直接输出a[i].time就好

否则活不下去就一个不叠 全吃了 也就是dp[0]

记得物品按时间排序

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 1000000007
using namespace std;
typedef long long ll;
typedef double D;
#define eps 1e-8
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
//head
const int N = 1005;
int n,V;

struct node {
    int h,t,x;
    bool operator < (const node &o) const {
        return x < o.x;
    }
}a[N];

int dp[N];
int main() {
    V = read(),n = read();
    rep(i,1,n) a[i].x = read(),a[i].t = read(),a[i].h = read();
    sort(a+1,a+n+1);
    dp[0] = 10;
    rep(i,1,n) {
        per(j,V,0) {
            if(dp[j] < a[i].x) continue;
            if(j + a[i].h >= V) {
                printf("%d\n",a[i].x);
                return 0;
            }
            dp[j + a[i].h] = max(dp[j + a[i].h],dp[j]);
            dp[j] += a[i].t;
        }
    }
    printf("%d\n",dp[0]);
    return 0;
}