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  • LUOGU P1779 魔鬼杀手_NOI导刊2010提高(03)

    传送门

    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;
    }
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  • 原文地址:https://www.cnblogs.com/sdfzsyq/p/9676863.html
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