解题思路
背包,首先先用aoe都打残然后单伤补刀,用f[i]表示AOE打了i的伤害的最小花费,g[i]表示单伤打了i的伤害的最小花费。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdio>
#include<cstring>
#define int long long
using namespace std;
const int MAXN = 105;
inline int rd(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)) {f=ch=='-'?0:1;ch=getchar();}
while(isdigit(ch)) {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return f?x:-x;
}
int f[100005],g[100005],ans=0x3f3f3f3f;
int n,m,val[MAXN],cost[MAXN],hp[MAXN];
bool al[MAXN];
signed main(){
n=rd();memset(f,0x3f,sizeof(f));
memset(g,0x3f,sizeof(g));
for(register int i=1;i<=n;i++) hp[i]=rd();
m=rd();char c[MAXN],s[MAXN];
for(register int i=1;i<=m;i++){
scanf("%s",c+1);cost[i]=rd();
scanf("%s",s+1);if(s[1]=='A') al[i]=1;
val[i]=rd();
if(cost[i]==0 && val[i]>0) {cout<<0<<endl;return 0;}
if(val[i]>100000) val[i]=100000;
}f[0]=0;g[0]=0;
for(register int i=1;i<=m;i++){
if(al[i]){
for(register int j=val[i];j<=100000;j++)
f[j]=min(f[j],f[j-val[i]]+cost[i]);
}
else {
for(register int j=val[i];j<=100000;j++)
g[j]=min(g[j],g[j-val[i]]+cost[i]);
}
}
for(register int i=99999;i>=0;i--)
if(g[i]>g[i+1]) g[i]=g[i+1]; //打了i+1的伤害就一定打了i的伤害
for(register int i=0;i<=100000;i++) {
int now=f[i];if(now>ans) continue;
for(register int j=1;j<=n;j++){
if(hp[j]>i) now+=g[hp[j]-i];
if(now>ans) break;
}
ans=min(ans,now);
}cout<<ans<<endl;
}