这题阶段性比较明显,可以事实上气力越大的乌龟放在越下面越有优势, So,阶段就是按照气力排序
状态: dp[i][j] 表示 前i个乌龟可以堆积成高度为j个最小重量
状态转移: dp[i][j] = min(dp[i][j-1], dp[i-1][j-1]+w[i](s[i] >= w[i]+dp[i-1][j-1]));
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define MAXN 5700
#define INF 0x3f3f3f3f
typedef struct NODE_ {
int w, s;
}NODE;
NODE turtle[MAXN];
int dp[MAXN][MAXN];
int cmp(const NODE &a, const NODE &b)
{
if( a.s == b.s ) {
return a.w < b.w;
}
return a.s < b.s;
}
int dynamic_pro(const int &n)
{
sort(turtle+1, turtle+n+1, cmp);
for(int i = 0; i <= n; i ++) {
dp[0][i] = dp[i][0] = 0;
}
for(int i = 1; i <= n; i ++) {
for(int j = 1; j <= i; j ++) {
dp[i][j] = (!dp[i-1][j])? INF : dp[i-1][j];
if( turtle[i].s >= dp[i-1][j-1]+turtle[i].w ) {
dp[i][j] = min(dp[i][j], dp[i-1][j-1]+turtle[i].w);
}
}
}
for(int i = n; i >= 1; i --) {
if( INF > dp[n][i] ) {
return i;
}
}
}
int main(int argc, char const *argv[])
{
#ifndef ONLINE_JUDGE
freopen("test.in", "r", stdin);
#endif
int idx(1);
while( ~scanf("%d %d", &turtle[idx].w, &turtle[idx].s) ) { idx ++; }
printf("%d\n", dynamic_pro(idx-1));
return 0;
}