百度之星资格赛1003：度度熊与邪恶大魔王

题面不kuai了，传送门

一道简单dp题，为了刷访问量写一篇吧

n巨大

防御力10

血量1000

10*1000=10⁴

技能1000

所以dp解决。

f[i][j]=min(f[i][j],f[i][max(0,j-p[l]+i)]+k[l]); (l=1...m)

自己理解就好了啦

ps.果真刷访问量神器

// It is made by XZZ
#include<cstdio>
#include<algorithm>
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
    rg int x=0,f=1;rg char ch=getchar();
    while(ch<'0'||ch>'9')f=ch=='-'?-1:f,ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
ll f[11][1001],k[1001],p[1001],a[100001],b[100001];
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)==2){
        rep(i,1,n)a[i]=gi(),b[i]=gi();
        rep(i,1,m)k[i]=gi(),p[i]=gi();
        rep(i,0,10)rep(j,1,1000)f[i][j]=1000000000000000ll;
        rep(i,0,10)rep(j,1,1000)rep(l,1,m)if(p[l]>i)
            f[i][j]=min(f[i][j],f[i][max(0ll,j-p[l]+i)]+k[l]);
        ll ans=0;
        rep(i,1,n)if(f[b[i]][a[i]]==1000000000000000ll)ans=-1;
        if(ans!=-1)rep(i,1,n)ans+=f[b[i]][a[i]];
        printf("%I64d\n",ans);
    }
    return 0;
}