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ZOJ3623 Battle Ships （完全背包）

編輯：C++入門知識

ZOJ3623 Battle Ships （完全背包）

Battle Ships

Time Limit: 2 Seconds Memory Limit: 65536 KB

Battle Ships is a new game which is similar to Star Craft. In this game, the enemy builds a defense tower, which has L longevity. The player has a military factory, which can produce N kinds of battle ships. The factory takes t_i seconds to produce the i-th battle ship and this battle ship can make the tower loss l_i longevity every second when it has been produced. If the longevity of the tower lower than or equal to 0, the player wins. Notice that at each time, the factory can choose only one kind of battle ships to produce or do nothing. And producing more than one battle ships of the same kind is acceptable.

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

Sample Output

2
4
5

題意：有n種船給出造出每種船的時間和該船一秒可以打多少滴血，給出L總血量，問打到L滴血最少需要多少秒。

設dp[i] 代表第i秒最多可以打掉多少滴血那麼答案就是從第0秒開始枚舉找到第一個大於等於L的就結束。

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

#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
//#include 
#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;
}