POJ 3265 DP
2016-10-14 00:24
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思路:
f[i][j]表示前i天能做j道题 (是做 不是做完)
if(f[i-1][k])
if(suma[j]-suma[k]+g[i-1][k]<=n)
f[i][j]=1,g[i][j]=min(g[i][j],sumb[j]-sumb[k]);
g[i][j]是转移的代价
当g[i][p]
//By SiriusRen #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int n,p,A[333],B[333],suma[333],sumb[333]; int g[333][333],f[333][333]; int main(){ scanf("%d%d",&n,&p); for(int i=1;i<=p;i++){ scanf("%d%d",&A[i],&B[i]); suma[i]=suma[i-1]+A[i]; sumb[i]=sumb[i-1]+B[i]; } memset(g,0x3f,sizeof(g)); for(int i=0;i<=p;i++)g[i][0]=0,g[0][i]=0; f[0][0]=1; for(int i=1;;i++){ for(int j=1;j<=p;j++) for(int k=0;k<=j;k++) if(f[i-1][k]) if(suma[j]-suma[k]+g[i-1][k]<=n) f[i][j]=1,g[i][j]=min(g[i][j],sumb[j]-sumb[k]); if(g[i][p]<n){printf("%d\n",i+2);return 0;} } }
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