前言

$Todayisthepenultimateday$

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成績(jī)

$Rank$是有算別人的

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

$3$	$2017zyc$	$250$	$90$	$100$	$60$
$10$	$2017xxy$	$220$	$20$	$100$	$100$
$21$	$2017lrz$	$180$	$20$	$100$	$60$
$21$	$2017wyc$	$180$	$30$	$50$	$100$
$54$	$2017xjq$	$110$	$0$	$50$	$60$
$72$	$2017hzb$	$80$	$0$	$20$	$60$
$81$	$2017lw$	$70$	$0$	$10$	$60$
$88$	$2017hjq$	$60$	$0$	$0$	$60$

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正題

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$T1:jzoj4252?$QYQ的圖【$dfs$】

博客鏈接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86702752

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$T2:jzoj4248?n$染色【數(shù)學(xué),快速冪】

博客鏈接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86703758

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$T3:jzoj4249?$游戲【貪心】

博客鏈接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86703879

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$someofcode$

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T1 30分code

#include<cstdio> #include<algorithm> using namespace std; const int N=60; int tot,in[N],ans,w[N],flag[N],n,maxs,m; bool v[N],a[N][N],zh[N]; void dfs(int x) {in[++tot]=x;v[x]=1;ans+=w[x];for(int i=1;i<=n;i++)if(a[x][i]&&!v[i])dfs(i); } void get_ans(int dep,int z) {if(dep>tot){maxs=max(maxs,z);return;}get_ans(dep+1,z);if(!flag[dep]&&!zh[in[dep]]){for(int i=1;i<=tot;i++)if(a[in[dep]][in[i]])flag[i]++;get_ans(dep+1,z+w[in[dep]]);for(int i=1;i<=tot;i++)if(a[in[dep]][in[i]])flag[i]--;} } int main() {freopen("graph.in","r",stdin);freopen("graph.out","w",stdout);scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)scanf("%d",&w[i]);for(int i=1;i<=m;i++){int x,y;scanf("%d%d",&x,&y);if(x==y) zh[x]=1;a[x][y]=a[y][x]=1;}for(int i=1;i<=n;i++){maxs=0;tot=0;dfs(i);get_ans(1,0);ans-=maxs;}printf("%d",ans); }

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T2 50分code

#include<cstdio> #define ll long long using namespace std; const ll XJQ=1e9+7; ll n,m,ans,f[100010][15]; int main() {freopen("color.in","r",stdin);freopen("color.out","w",stdout);scanf("%lld%lld",&n,&m);f[1][1]=1;for(ll i=2;i<=n+1;i++)for(ll j=1;j<=m;j++)for(ll k=1;k<=m;k++)if(k!=j)(f[i][j]+=f[i-1][k])%=XJQ;printf("%lld",f[n+1][1]*m%XJQ); }

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總結(jié)

爆炸了

T1不會(huì)，去寫(xiě)T2，然后寫(xiě)了個(gè)錯(cuò)的。之后寫(xiě)T3，開(kāi)始寫(xiě)dp，后來(lái)發(fā)現(xiàn)可以貪心，切掉。然后T1寫(xiě)了個(gè)暴力(我還證明了暴力可過(guò))，結(jié)果寫(xiě)掛了，炸掉30。然后T2找了半天規(guī)律，最后二分鐘開(kāi)始寫(xiě)，結(jié)果時(shí)間不夠，只有之前那個(gè)50分codeQVQ

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尾聲

雷大爺險(xiǎn)些$Rank1$

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