TYVJ 1057 金明的预算方案 解题报告

　　这题好像是某届NOIP的原题，我用的简单DP，因为主件最多两个子件，所以没什么考虑的，多判断几次，但是这代码太猥琐了，写的我一头的汗。不过还好，一次AC！

================================华丽的分割线 ================================

#include <stdio.h>
#include <stdlib.h>
struct thing{
	int p, i;
	int len;
	int sub[2];
}thing[60];
int end;
int f[32001];

int main(int argc, char **argv)
{
	int i, j;
	int n, m;
	int a, b, c;
	scanf("%d%d", &m, &n);
	for(i = 0; i < n; i++){
		scanf("%d%d%d", &a, &b, &c);
		thing[end].p = a;
		thing[end].i = b;
		if(c != 0){
			thing[c - 1].sub[thing[c - 1].len++] = i;
			//说明它不是主件
			thing[end].len = -1;
		}
		end++;
	}
	for(i = 0; i < n; i++){
		if(thing[i].len == -1){
			continue;
		}
		for(j = m; j >= thing[i].p; j--){
			if(f[j] < f[j - thing[i].p] + (thing[i].p * thing[i].i)){
				f[j] = f[j - thing[i].p] + (thing[i].p * thing[i].i);
			}
			if(thing[i].len >= 1){
				if(j - thing[i].p - thing[thing[i].sub[0]].p >= 0 && f[j] < f[j - thing[i].p - thing[thing[i].sub[0]].p] +
						(thing[i].p * thing[i].i + thing[thing[i].sub[0]].p * thing[thing[i].sub[0]].i)){
					f[j] = f[j - thing[i].p - thing[thing[i].sub[0]].p] +
							(thing[i].p * thing[i].i + thing[thing[i].sub[0]].p * thing[thing[i].sub[0]].i);
				}
				if(thing[i].len == 2){
					if(j - thing[i].p - thing[thing[i].sub[1]].p >= 0 && f[j] < f[j - thing[i].p - thing[thing[i].sub[1]].p] +
							(thing[i].p * thing[i].i + thing[thing[i].sub[1]].p * thing[thing[i].sub[1]].i)){
						f[j] = f[j - thing[i].p - thing[thing[i].sub[1]].p] +
								(thing[i].p * thing[i].i + thing[thing[i].sub[1]].p * thing[thing[i].sub[1]].i);
					}

					if(j - thing[i].p - thing[thing[i].sub[0]].p - thing[thing[i].sub[1]].p >= 0
							&& f[j] < f[j - thing[i].p - thing[thing[i].sub[0]].p - thing[thing[i].sub[1]].p] +
							(thing[i].p * thing[i].i + thing[thing[i].sub[1]].p * thing[thing[i].sub[1]].i + 
							thing[thing[i].sub[0]].p * thing[thing[i].sub[0]].i)){
						f[j] = f[j - thing[i].p - thing[thing[i].sub[0]].p - thing[thing[i].sub[1]].p] +
							(thing[i].p * thing[i].i + thing[thing[i].sub[1]].p * thing[thing[i].sub[1]].i + 
							thing[thing[i].sub[0]].p * thing[thing[i].sub[0]].i);
					}

				}
			}
		}
	}
	printf("%d\n", f[m]);
	return 0;
}