ZOJ3623 Battle Ships (完全揹包)
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 ti seconds to produce the i
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
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
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秒開始列舉 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 <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;
}