大容量背包逆向思维
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std;
const int maxn = 507;
int dp[maxn + 1];
int w[maxn], v[maxn];
int main()
{
int T;
scanf("%d", &T);
while (T--) {
int n, B, V = 0;
scanf("%d%d", &n, &B);
for (int i = 0; i < n; ++i) { //重量不超过B, sum (v) <= 5000
scanf("%d%d", w + i, v + i);
V += v[i];
}
memset(dp, 0x3f, sizeof dp);
dp[0] = 0; //就错在dp这
for (int i = 0; i < n; ++i) {
for (int j = V; j >= v[i]; --j) { //dp[j]表示装价值为j时所需的最小容量
if (dp[j - v[i]] + w[i] <= B) { //所需容量尽量小
dp[j] = min(dp[j], dp[j - v[i]] + w[i]);
}
}
}
int ans = -1;
for (int j = V; j >= 0; --j) {
printf("dp[%d] = %d\n", j, dp[j]);
if (dp[j] <= B) {
ans = j;
break;
}
}
printf("%d\n", ans); //价值， dp[j]为重量，dp[j]的价值为j
}
return 0;
}
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