ZOJ - 3623 - Battle Ships （01背包变形）

Battle Ships

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

1 11 12 101 12 53 1001 103 2010 100

Sample Output

245

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Author: FU, Yujun
Contest: ZOJ Monthly, July 2012
Submit    Status

思路：将时间翻转过来，让时间作为背包的容量来进行动态规划，，，

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>#define INF 0x3fffffffusing namespace std;int t[105], l[105];int dp[505];//dp[i]表示在i这个时间以内所造成的最大伤害 int main() {int N, L;while(scanf("%d %d", &N, &L) != EOF) {for(int i = 1; i <= N; i++) {scanf("%d%d", &t[i], &l[i]);}memset(dp, 0, sizeof(dp));for(int i = 0; i <= 335; i++) {for(int j = 1; j <= N; j++) {dp[i + t[j]] = max(dp[i + t[j]], dp[i] + i * l[j] );}} int ans = INF;for(int i = 0; i <= 335; i++) {if(dp[i] >= L && ans > i) ans = i;}printf("%d\n", ans);}return 0;}