hdu 5777 domino（贪心）

多米诺博弈最小化高度和

最新推荐文章于 2019-12-05 18:07:23 发布

原创最新推荐文章于 2019-12-05 18:07:23 发布 · 743 阅读

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探讨一种多米诺博弈策略，通过限定推倒次数和优化高度布局，实现所有多米诺骨牌总高度最小化的问题及解决方案。

domino

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 179 Accepted Submission(s): 105

Problem Description

Little White plays a game.There are n pieces of dominoes on the table in a row. He can choose a domino which hasn't fall down for at most k times, let it fall to the left or right. When a domino is toppled, it will knock down the erect domino. On the assumption that all of the tiles are fallen in the end, he can set the height of all dominoes, but he wants to minimize the sum of all dominoes height. The height of every domino is an integer and at least 1.

Input

The first line of input is an integer T (

1≤T≤10 )
There are two lines of each test case.
The first line has two integer n and k, respectively domino number and the number of opportunities.(

2≤k,n≤100000 )
The second line has n - 1 integers, the distance of adjacent domino d,

1≤d≤100000

Output

For each testcase, output of a line, the smallest sum of all dominoes height

Sample Input

Sample Output

Source

BestCoder Round #85

题意：地上有n个多米诺，最多推倒k次，前一个多米诺的高度>=两者距离+1可以推倒后面的，现在问n个多米诺最小高度之和是多少

思路：贪心即可。把前k个最大的路径断掉，其他的高度等于路径+1即可

代码：

#include<cstring>
#include <algorithm>
#include <math.h>
#include <queue>
#include <stdio.h>
using namespace std;
const int N = 100005;
int a[N];
int cmp(int a,int b){
    return a>b;
}
int main(){
    int tcase,n,k;
    scanf("%d",&tcase);
    while(tcase--){
        scanf("%d%d",&n,&k);
        long long sum = 0;
        for(int i=1;i<n;i++){
            scanf("%d",&a[i]);
            sum=sum+a[i]+1;
        }
        if(n<=k){
            printf("%d\n",n);
            continue;
        }
        sort(a+1,a+n,cmp);
        for(int i=1;i<k;i++){
            sum-=a[i];
        }
        printf("%I64d\n",sum+1);
    }
    return 0;
}