HDOJ  4310   Hero

题目：http://acm.hdu.edu.cn/showproblem.php?pid=4310

题目解读：可以算出每一位敌人的单位血的伤害量，进行排序，先干掉最大的，，，，

#include <stdio.h>
#include <stdlib.h>
struct hero
{
    intDPS,HP;
    doublelife;
}aaa[25];
int cmp(const void *a ,const void *b)
{
    struct hero*c=(hero *)a;
    struct hero*d=(hero *)b;
    returnc->life < d->life ? 1:-1;           //double的排序返回的一定要注意，不能直接减
}
int main()
{
    inti,j,n,sum,hurt;
   while(scanf("%d",&n)!=EOF)
    {
       sum=0;
       hurt=0;
       for(i=0;i<n;i++)
       {
           scanf("%d%d",&aaa[i].DPS,&aaa[i].HP);
           aaa[i].life=1.0*aaa[i].DPS/aaa[i].HP;
           sum+=aaa[i].DPS;
       }
       qsort(aaa,n,sizeof(aaa[0]),cmp);
       for(i=0;i<n;i++)
       {
           for(j=1;j<=aaa[i].HP;j++)
              hurt+=sum;
           sum-=aaa[i].DPS;
       }
       printf("%d\n",hurt);
    }
    return0;
}
还有一种方法是：

状态压缩dp，用dp[mask]表示杀死mask集合的敌人时，这些敌人造成的最小hp消耗。

有转移方程dp[mask] = min{dp[mask - {i}] + hp_sum[mask] * dps[i], forall i in mask}

 

#include <cstdio>
#include <cstring>

const int MAXN = 20;
struct SS {int dps, hp;} a[MAXN];
int sum[1<<MAXN],dp[1<<MAXN];
int n;

int main() {
  // freopen("data.in","r",stdin);
  // freopen("data.out","w",stdout);
    while(scanf("%d",&n) != EOF) {
       for (int i = 0 ; i < n ; i++)
           scanf("%d%d",&a[i].dps,&a[i].hp);
       memset(sum, 0, sizeof(sum));
       for (int mask = 0 ; mask <(1<<n) ; ++mask)
           for (int i = 0 ; i < n ; i++)
               if (mask&(1<<i))

                     sum[mask] += a[i].hp;
       memset(dp, 0x3f, sizeof(dp));
       dp[0] = 0;
       for (int mask = 0 ; mask <(1<<n) ; ++mask) {
           for (int i = 0 ; i < n ; i++) {
               if (mask & (1<<i))continue;
               int tmask = mask + (1<<i);
               int cost = dp[mask] + sum[tmask] * a[i].dps;
               if (cost < dp[tmask]) dp[tmask] = cost;
           }
       }
       printf("%d\n", dp[(1<<n)-1]);
    }
    return0;
}