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
1 // It is made by XZZ 2 #include<cstdio> 3 #include<algorithm> 4 using namespace std; 5 #define rep(a,b,c) for(rg int a=b;a<=c;a++) 6 #define drep(a,b,c) for(rg int a=b;a>=c;a--) 7 #define erep(a,b) for(rg int a=fir[b];a;a=nxt[a]) 8 #define il inline 9 #define rg register 10 #define vd void 11 typedef long long ll; 12 il int gi(){ 13 rg int x=0,f=1;rg char ch=getchar(); 14 while(ch<'0'||ch>'9')f=ch=='-'?-1:f,ch=getchar(); 15 while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar(); 16 return x*f; 17 } 18 ll f[11][1001],k[1001],p[1001],a[100001],b[100001]; 19 int main(){ 20 int n,m; 21 while(scanf("%d%d",&n,&m)==2){ 22 rep(i,1,n)a[i]=gi(),b[i]=gi(); 23 rep(i,1,m)k[i]=gi(),p[i]=gi(); 24 rep(i,0,10)rep(j,1,1000)f[i][j]=1000000000000000ll; 25 rep(i,0,10)rep(j,1,1000)rep(l,1,m)if(p[l]>i) 26 f[i][j]=min(f[i][j],f[i][max(0ll,j-p[l]+i)]+k[l]); 27 ll ans=0; 28 rep(i,1,n)if(f[b[i]][a[i]]==1000000000000000ll)ans=-1; 29 if(ans!=-1)rep(i,1,n)ans+=f[b[i]][a[i]]; 30 printf("%I64d ",ans); 31 } 32 return 0; 33 }