洛谷 1156 垃圾陷阱

作为一道蓝色的DP神题，它成功把我卡了两个小时。

这个题对我来说确实有些困难，本人太菜了，对DP毫无感觉。

后来发现这是个背包，要么吃要么放，前提是饿不死。

于是：

AC代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
inline int read()
{
    int x = 0;
    int k = 1;
    char c = getchar();
    while (!isdigit(c))
        if (c == '-') k = -1, c = getchar();
        else c = getchar();
    while (isdigit(c))
        x = (x << 1) + (x << 3) + (c ^ 48),
        c = getchar();
    return k * x;
}
int m, n, dp[105][3000];
struct node
{
    int t, f, h;
    bool operator < (const node & a) const
        { return t < a.t; }
}rbs[102];
int main()
{
    m = read();
    n = read();
    memset(dp, -1, sizeof(dp));
    for (int i = 1; i <= n; ++i) 
        rbs[i].t = read(), rbs[i].f = read(), rbs[i].h = read();
    sort(rbs + 1, rbs + n + 1);
    dp[0][0] = 10;
    for (int i = 0; i <= n; ++i)
        for (int j = 0; j <= m; ++j)
        {
            if (dp[i][j] < 0) continue;
            if (dp[i][j] >= rbs[i + 1].t - rbs[i].t && j + rbs[i + 1].h >= m)
            {
                printf("%d", rbs[i + 1].t);
                return 0;
            }
            if (dp[i][j] >= rbs[i + 1].t - rbs[i].t)
                dp[i + 1][j + rbs[i + 1].h] = dp[i][j] - rbs[i + 1].t + rbs[i].t,
                dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + rbs[i + 1].f - rbs[i + 1].t + rbs[i].t);
        }
    int sum = 0, mogic = 10;
    for (int i = 1; i <= n; ++i)
        if (mogic >= rbs[i].t - rbs[i - 1].t)
            mogic -= rbs[i].t - rbs[i - 1].t, mogic += rbs[i].f, sum += rbs[i].t - rbs[i - 1].t;
        else 
        {
            printf("%d", sum + mogic);
            return 0;
        }
    printf("%d", sum + mogic);
    return 0;
}