ZOJ 3689 Digging(貪心+dp)
When it comes to the Maya Civilization, we can quickly remind of a term called the end of the world. It‘s not difficult to understand why we choose to believe the prophecy (or we just assume it is true to entertain ourselves) if you know the other prophecies appeared in the Maya Calendar
The ancient civilization, such as Old Babylonianhas, Ancient Egypt and etc, some features in common. One of them is the tomb because of the influence of the religion. At that time, the symbol of the tomb is the pyramid. Many of these structures featured a top platform upon which a smaller dedicatory building was constructed, associated with a particular Maya deity
Now there are N coffin chambers in the pyramid waiting for building and the ruler has recruited some workers to work for T days. It takes ti days to complete the ith coffin chamber. The size of the ith
At the beginning, there are T days left. If they start the last work at the time t and the finishing time t-ti < 0, they will not get the last pay.
Input
There are few test cases.
The first line contains N, T (1 ≤ N ≤ 3000,1 ≤ T ≤ 10000), indicating there are N coffin chambers to be built, and there are T days for workers working. Next N lines contains ti, si (1 ≤ti, si ≤ 500).
All numbers are integers and the answer will not exceed 2^31-1.
Output
For each test case, output an integer in a single line indicating the maxminal units of gold the workers will get.
Sample Input
3 10 3 4 1 2 2 1
Sample Output
62
Hint
題意:n個任務,剩余t的時間,完畢每一個任務會消耗時間ti,同一時候也會得到t*si的價值,求怎麽安排任務得到的價值最大題解:一開始僅僅想用背包,出不來例子。排個序後就能夠了。
至於為什麽。我也不知道,(看來還是沒參透dp的精髓。
。),想來大概是背包得先背性價比高的來保證貪心的思想吧。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define ll long long using namespace std; const int M=10010; const int N=3030; int n,T; ll dp[M]; struct node { int t,v; } a[N]; bool cmp(node a,node b) { return a.v*b.t<a.t*b.v; } int main() { // freopen("test.in","r",stdin); while(~scanf("%d%d",&n,&T)) { for(int i=0; i<n; i++) scanf("%d%d",&a[i].t,&a[i].v); sort(a,a+n,cmp); memset(dp,0,sizeof dp); for(int i=0; i<n; i++) { for(int j=T; j>=a[i].t; j--) { dp[j]=max(dp[j],dp[j-a[i].t]+a[i].v*j); } } printf("%lld\n",dp[T]); } return 0; }
ZOJ 3689 Digging(貪心+dp)