hdu2831Plants VS Zombies (贪心)

Plants VS Zombies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 623    Accepted Submission(s): 264
Special Judge

Problem Description

Plants VS Zombies is a popular game in the TJU ACM Team now. However, RoBa is not very good at this game, so he wants you to write a program to help him.

We take a simplified version of this game. There are N rows in your backyard, and there are N zombies want to walk through your yard to eat your brain, and exactly one zombie in each row. The only method to defense the zombies is to plant “peashooter” on the other side of the yard. However, you need T seconds to plant one peashooter. Once the peashooter is planted, it will begin to shoot at the zombies in its row. The zombies have different speed Vi and defense Di, which means that, the zombie in the i-th row need Vi second the walk though the yard, and the i-th zombie will be killed after Di second of shooting. Please note you can only plant one peashooter in each row.

Unfortunately, in some cases, for the zombies are too fast and/or too strong and/or your planting is too slow, you cannot kill all the zombies in time, so the zombies will eat your brains.

 

Input

There are several test cases in the input. The first line of each test case contains two integers N (1 <= N <= 100) and T, indicating the number of rows and the time for planting one peashooter. Then N lines follow, each of which contains two integers Vi and Di.

 

Output

Output one line for each test case. If you can kill all the zombies, output N numbers separated by a space, that is, a permutation of 1,2,…N, indicating the order of your planting. If you cannot defend your yard, output “The zombies eat your brains!” in one line.

 

Sample Input

3 1020 1040 1030 103 1020 1040 1130 10

 

Sample Output

1 3 2The zombies eat your brains!
题目大意：在 n 列中种下豌豆射手来阻挡每行的僵尸，且已知豌豆是每 t 秒种一棵，第i列的僵尸进入院子需要 v[i]秒，杀死需要 d[i]秒。
问能否阻挡所有僵尸进入院子，若能，打印出僵尸被消灭的列数的顺序。若否打印“The zombies eat your brains!”。
解题思路：经典的贪心问题，由题意知，僵尸进入院子所需时间v[i]较小的，而杀死他所需时间d[i]较大的应先被消灭，v[i]较大，而d[i]较小的可以放到后面再消灭。
即v[i]-d[i]的差值较小的那列应先种上豌豆，所以需要对v[i]-d[i]进行升序排列。同时，若k为消灭前i-1列僵尸所用时间，则当k+d[i]<=v[i]时，第i列僵尸才能被消灭，否则，第i列就不能被消灭。
代码：
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{int v,d;int r;}ans[120];int cmp(node x,node y){return (x.v-x.d)<(y.v-y.d);}int main(){int n,t;while(scanf("%d%d",&n,&t)!=EOF){memset(ans,0,sizeof(ans));for(int i=1;i<=n;i++){scanf("%d%d",&ans[i].v,&ans[i].d);ans[i].r=i;}sort(ans+1,ans+n+1,cmp);int flag=1;int k=t;for(int i=1;i<=n;i++){if(ans[i].d+k<=ans[i].v){k+=t;}else{flag=0;break;}}if(flag){printf("%d",ans[1].r);for(int i=2;i<=n;i++) printf(" %d",ans[i].r);printf("\n"); }else printf("The zombies eat your brains!\n");}return 0;}