前言

$Rank1$耶

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成績

$Rank$是有算別人的

$Rank$$Person$$Score$$A$$B$$C$

$1$	$2017myself$	$280$	$100$	$80$	$100$
$3$	$2017xjq$	$200$	$100$	$0$	$100$
$12$	$2017hjq$	$150$	$100$	$0$	$50$
$30$	$2017zyc$	$100$	$100$	$0$	$0$
$30$	$2017xxy$	$100$	$100$	$0$	$0$
$30$	$2017lrz$	$100$	$100$	$0$	$0$
$30$	$2017hzb$	$100$	$100$	$0$	$0$
$30$	$2017lw$	$100$	$100$	$0$	$0$

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正題

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$T1:jzoj4244?yi$【貪心】

博客鏈接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86689041

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$T2:jzoj4245?er$【$dp,$貪心】

博客鏈接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86689149

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$T3:jzoj4246?san$【最短路$,SPFA,DAGdp$】

博客鏈接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86689203

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$someofcode$

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T2 80分code

#include<cstdio> #include<algorithm> #include<cstring> #include<cmath> using namespace std; const int N=110; int n,m,k,add[N],sum[N],cnt1,cnt2,mul[N]; int fff,f[N][N*2000],maxs[N],a1,a2; double ans,summ[N]; bool cmp(int x,int y) {return x>y;} int main() {scanf("%d%d%d",&n,&m,&k);scanf("%d",&a1);if(n==2) scanf("%d",&a2);for(int i=1;i<=m;i++){int x,t;scanf("%d%d",&t,&x);if(t==1) fff=x;if(t==2) add[++cnt1]=x;if(t==3) mul[++cnt2]=x;}sort(add+1,add+1+cnt1,cmp);sort(mul+1,mul+1+cnt2,cmp);sum[0]=a1+a2;for(int i=1;i<=max(cnt1,cnt2);i++){summ[i]=summ[i-1]+log(mul[i]);sum[i]=sum[i-1]+add[i];}memset(f,0xcf,sizeof(f));f[0][a1]=a1*a2;for(int i=0;i<=cnt1;i++){for(int j=a1;j<=sum[i];j++){f[i+1][j]=max(f[i][j]+add[i+1]*j,f[i+1][j]);f[i+1][j+add[i+1]]=max(f[i][j]+add[i+1]*(sum[i]-j),f[i+1][j+add[i+1]]);maxs[i]=max(maxs[i],f[i][j]);}}for(int u=0;u<=k;u++){int mu=k-u;ans=max(ans,log(maxs[u])+summ[mu]);}if(fff){k--;sum[0]=fff+a2;for(int i=1;i<=cnt1;i++)sum[i]=sum[i-1]+add[i];memset(f,0xcf,sizeof(f));memset(maxs,0,sizeof(maxs));f[0][fff]=fff*a2;for(int i=0;i<=cnt1;i++){for(int j=fff;j<=sum[i];j++){f[i+1][j]=max(f[i][j]+add[i+1]*j,f[i+1][j]);f[i+1][j+add[i+1]]=max(f[i][j]+add[i+1]*(sum[i]-j),f[i+1][j+add[i+1]]);maxs[i]=max(maxs[i],f[i][j]);}}for(int u=0;u<=k;u++){int mu=k-u;ans=max(ans,log(maxs[u])+summ[mu]);}sum[0]=a1+fff;for(int i=1;i<=cnt1;i++)sum[i]=sum[i-1]+add[i];memset(f,0xcf,sizeof(f));memset(maxs,0,sizeof(maxs));f[0][a1]=a1*fff;for(int i=0;i<=cnt1;i++){for(int j=a1;j<=sum[i];j++){f[i+1][j]=max(f[i][j]+add[i+1]*j,f[i+1][j]);f[i+1][j+add[i+1]]=max(f[i][j]+add[i+1]*(sum[i]-j),f[i+1][j+add[i+1]]);maxs[i]=max(maxs[i],f[i][j]);}}for(int u=0;u<=k;u++){int mu=k-u;ans=max(ans,log(maxs[u])+summ[mu]);}}printf("%.3lf",ans); }

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T2 60分code

#include<cstdio> #include<algorithm> #include<cmath> using namespace std; const int N=110; int n,m,k,a1,t,x,add[N],mul[N],cnt1,cnt2,f,a2; double ans,sum[N]; bool cmp(int x,int y) {return x>y;} void check_nf(int A1,int A2) {for(int u=0;u<=min(k,cnt1);u++){int a1=A1,a2=A2;for(int i=1;i<=u;i++)if(n==1||a1<a2) a1+=add[i];else a2+=add[i];//printf("%d %d\n",a1,a2);ans=max(ans,log(a1)+log(a2)+sum[min(k-u,cnt2)]);} } void check_f(int A1,int A2) {k--;if(A1<A2) A1=f;else A2=f;for(int u=0;u<=min(k,cnt1);u++){int a1=A1,a2=A2;for(int i=1;i<=u;i++)if(n==1||a1<a2) a1+=add[i];else a2+=add[i];ans=max(ans,log(a1)+log(a2)+sum[min(k-u,cnt2)]);} } int main() {//freopen("data.in","r",stdin);//freopen("data.ans","w",stdout);scanf("%d%d%d",&n,&m,&k);scanf("%d",&a1);if(n==2) scanf("%d",&a2);for(int i=1;i<=m;i++){int t,x;scanf("%d%d",&t,&x);if(t==1) f=x;else if(t==2)add[++cnt1]=x;else if(t==3)mul[++cnt2]=x;}sort(add+1,add+1+cnt1,cmp);sort(mul+1,mul+1+cnt2,cmp);for(int i=1;i<=cnt2;i++)sum[i]=sum[i-1]+log(mul[i]);check_nf(a1,a2);if(f) check_f(a1,a2);printf("%0.3lf",ans);}

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總結

T1看完題之前以為是之前做過的一道惡心的dp，結果發現是水題，切掉。然后看T2，開始不會自然對數，結果發現就是log，然后寫了個貪心先。之后看T3，先寫了一個多源最短路，然后發現之后不會，之后仔細看一下題目，發現其實只有最短路上的邊有用，這樣就是有向無環圖了，切掉。
之后T2怕貪心是錯的，寫個dp，調了半天。只有80。考完后發現是沒有特判n=1的情況。

看題預估分$(100+20+30=150)$

做題預估分$(100+50+50=200)$

實際分數$(100+80+100=280)$

$Rank1?$還可以耶

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