前言

昨天：
$Md$今天还真爆零了

顺便$\%\%\%ZZYRank1$

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成绩

$J$表示初中，$H$表示高中后面加的是几年级

$Rank$	$Person$	$Score$	$A$	$B$	$C$
$1$	$(H-1)ZZY$	$150$	$20$	$30$	$100$
$3$	$(H-1)WHF$	$120$	$20$	$0$	$100$
$36$	$(J-3)XXY$	$40$	$0$	$10$	$30$
$36$	$(J-3)ZYC$	$40$	$0$	$40$	$0$
$36$	$(J-3)LRZ$	$40$	$0$	$10$	$30$
$44$	$(J-3)HZB$	$30$	$0$	$0$	$30$
$64$	$(H-1)QYH$	$20$	$20$	$0$	$0$
$75$	$(J-3)LW$	$0$	$0$	$0$	$0$
$75$	$(H-2)HJW$	$0$	$0$	$0$	$0$
$75$	$(H-2)LZX$	$0$	$0$	$0$	$0$
$75$	$(J-3)WYC$	$0$	$0$	$0$	$0$

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正题

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$T1:jzoj3736-[NOI2014$模拟$7.11]$数学题$(math)$【计算几何】

链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/99098615

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$T2:jzoj3737-[NOI2014$模拟$7.11]$挖宝藏$(treasure)$【斯坦纳树$,SPFA,$状压】

链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/99102463

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$T3:jzoj3738-[NOI2014$模拟$7.11]$理想城市$(city)$【树$,$模型转换】

链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/99104932

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总结

一看$T1$没学过向量，$T2$好像很难，$T3$也很诡异。今天三题赶脚都不可做，有一种啥都不会的赶脚。发现$T2$好像有些可做就刚了一个小时多两个小时，然后瞬间爆炸没写出了。这时候$XJQ$告诉了我$T1$的向量运算，然后推了二十分钟多不会。然后看$T3$诶这个不是可以变成树形结构吗？？？然后就愉快刚了半天到比赛结束也没写暴力。

后来发现应该横竖各来一次。

ba说了，爆蛋

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$Someofcode$

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$T2WAcode$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define p(x,y) ((x-1)*m+y)
using namespace std;
const int N=11,dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};
struct node{int x,y,w;
}e[N*N];
int h,n,m,a[N][N][N],skr[N][N][N],f[N][N*N][N*N],num[N],K[N],fa[N],ans,cnt,zans,v[N*N];
int find(int x)
{return fa[x]==x?x:find(fa[x]);}
void unionn(int x,int y)
{int Fa=find(x),Fb=find(y);if(Fa==Fb) return;if(Fa<Fb) fa[Fb]=Fa;else fa[Fa]=Fb; 
}
bool cmp(node x,node y)
{return x.w<y.w;}
int main()
{freopen("treasure.in","r",stdin);freopen("treasure.out","w",stdout);scanf("%d%d%d",&h,&n,&m);for(int i=1;i<=h;i++)for(int j=1;j<=n;j++)for(int k=1;k<=m;k++)scanf("%d",&a[i][j][k]);memset(f,0x3f,sizeof(f));memset(skr,0x3f,sizeof(skr));memset(skr[0],0,sizeof(skr[0]));for(int z=1;z<=h;z++){for(int i=1;i<=n;i++)for(int j=1;j<=m;j++){fa[p(i,j)]=p(i,j);f[z][p(i,j)][p(i,j)]=0;for(int k=0;k<4;k++){int zx=i+dx[k],zy=j+dy[k];if(zx<1||zy<1||zx>n||zy>m) continue;f[z][p(i,j)][p(zx,zy)]=0;}v[p(i,j)]=a[z][i][j];}for(int k=1;k<=n*m;k++)for(int i=1;i<=n*m;i++)for(int j=1;j<=n*m;j++)f[z][i][j]=min(f[z][i][j],f[z][i][k]+f[z][k][j]+v[k]);scanf("%d",&K[z]);int maxs=0;for(int j=1;j<=K[z];j++){int x,y;scanf("%d%d",&x,&y);num[j]=p(x,y);maxs+=a[z][x][y];}K[z]++;for(int x1=1;x1<=n;x1++)for(int y1=1;y1<=m;y1++){ans=0;cnt=0;num[K[z]]=p(x1,y1);for(int i=1;i<K[z];i++)if(num[i]==p(x1,y1)) ans=-a[z][x1][y1];for(int i=1;i<=K[z];i++)fa[num[i]]=num[i];for(int i=1;i<=K[z];i++)for(int j=i+1;j<=K[z];j++)e[++cnt]=(node){num[i],num[j],f[z][num[i]][num[j]]};sort(e+1,e+1+cnt,cmp);for(int i=1;i<=cnt;i++){int Fa=find(e[i].x),Fb=find(e[i].y);if(Fa==Fb) continue;if(Fa<Fb) fa[Fb]=Fa;else fa[Fa]=Fb;ans+=e[i].w;}ans+=skr[z-1][x1][y1]+maxs+a[z][x1][y1];skr[z][x1][y1]=min(skr[z][x1][y1],ans);for(int x2=1;x2<=n;x2++)for(int y2=1;y2<=m;y2++)skr[z][x2][y2]=min(skr[z][x2][y2],ans+f[z][p(x1,y1)][p(x2,y2)]+a[z][x2][y2]); }if(K[z]==1) continue;ans=2147483647;for(int x1=1;x1<=n;x1++)for(int y1=1;y1<=m;y1++)ans=min(ans,skr[z][x1][y1]);zans=ans;}printf("%d",zans);
} 

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话说

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