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
 {
     int DPS,HP;
     double life;
 }aaa[25];
 int cmp(const void *a ,const void *b)
 {
     struct hero *c=(hero *)a;
     struct hero *d=(hero *)b;
     return c->life < d->life ? 1 : -1;            //double的排序返回的一定要注意，不能直接减
 }
 int main()
 {
     int i,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);
     }
     return 0;
 }
 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], for all i

#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++)
                                     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]);
     }
     return 0;
 }