uva 10154 Weights and Measures(01背包）

题目连接：10154 - Weights and Measures

题目大意：有若干只乌龟，每只乌龟有重量和负重两个属性，现在求这若干只乌龟最多可以叠多少层。

解题思路：01背包的变形， 以负重值作为背包的大小， 层数为所求dp值， 每次以min(当前可负重值-乌龟的重量, 乌龟的负重）考虑。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 10005;
int min(int a, int b) {	return a > b ? b : a; }

struct state {
    int w;
    int s;
}tmp[N];
int n, dp[N * 1000];

bool cmp (const state& a, const state& b) {
    return a.s > b.s;
}

void Init() {
    n = 0;
    memset(tmp, 0, sizeof(tmp));
    memset(dp, 0, sizeof(dp));
    while (scanf("%d%d", &tmp[n].w, &tmp[n].s) == 2)
	    n++;
    sort(tmp, tmp + n, cmp);
}

int solve() {
    int Max = tmp[0].s, t;
    for (int i = 0; i < n; i++) {
	for (int j = tmp[i].w; j <= Max; j++) {
	    if (dp[j]) {
		t = min(j - tmp[i].w, tmp[i].s);
		if (dp[t] < dp[j] + 1)
		    dp[t] = dp[j] + 1;
	    }
	}
	t = tmp[i].s - tmp[i].w;
	if (dp[t] == 0)
	    dp[t] = 1;
    }

    int cur = 0;
    for (int i = 0; i <= Max; i++) {
	if (cur < dp[i])    cur = dp[i];
    }
    return cur;
}

int main() {
    Init();

    printf("%d\n", solve());
    return 0;
}