ZOJ3623 Battle Ships （完全背包）

Battle Ships

Your job is to find out the minimum time the player should spend to win the game.

Input

There are multiple test cases.

The first line of each case contains two integers N(1 ≤ N ≤ 30) and
L(1 ≤ L ≤ 330), N is the number of the kinds of Battle Ships,
L is the longevity of the Defense Tower. Then the following N lines, each line contains two integers
t _i(1 ≤ t _i ≤ 20) and
l_i(1 ≤ l_i ≤ 330) indicating the produce time and the lethality of the i-th kind Battle Ships.

Output

Output one line for each test case. An integer indicating the minimum time the player should spend to win the game.

Sample Input

1 1
1 1
2 10
1 1
2 5
3 100
1 10
3 20
10 100

Sample Output

2
4
5

题意：有n种船 给出造出每种船的时间 和 该船一秒可以打多少滴血，给出L总血量，问打到L滴血最少需要多少秒。

设dp[i] 代表第i秒最多可以打掉多少滴血 那么答案就是从第0秒开始枚举 找到第一个大于等于L的就结束。

dp[j+t[i]] = max{ dp[i+t[i]], dp[j] + l[i]*j }

#include <algorithm>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <cstring>
#include <cstdlib>
#include <iostream>
//#include <bits/stdc++.h>
#define MAX 0x3f3f3f3f
#define N 100005
#define M 200005
#define mod 1000000007
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
typedef long long LL;
using namespace std;
const double pi = acos(-1.0);

int n, m;
int t[40], l[40], dp[40000];

int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d", &n, &m)) {

        for(int i = 0; i < n; i++) {
            scanf("%d%d", &t[i], &l[i]);
        }

        memset(dp, 0, sizeof(dp));

        for(int i = 0; i < n; i++) {
            for(int j = 0; j <= 340; j++) {
                dp[ t[i] + j ] = max(dp[ t[i] + j ], dp[j] + j*l[i]);
            }
        }
        for(int i = 0; i < 400; i++) {
            if(dp[i] >= m) {
                printf("%d\n", i);
                break;
            }
        }

    }

	return 0;
}