[Contest] THUSC 2016 解题报告

T1

想到了区间dp 没想到再加两维

f[i][j][down][up] 表示在区间[i,j]取数 还剩下[down,up]中的数的最小代价

转移f[i][j][down][up]=min{初始值,f[i][k][down][up]+f[k+1][j][down][up]}

其中 初始值 为[i,j]中去掉头尾连续的都处于值区间的数后取完的最小代价

取完的代价g[i][j]=min{f[i][j][down][up]+去掉剩下的数的代价}

代码么 有三份

单峰

<span style="font-family:Microsoft YaHei;font-size:14px;color:#000066;">#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;inline char nc(){static char buf[100000],*p1=buf,*p2=buf;if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }return *p1++;}inline void read(ll &x){char c=nc(),b=1;for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline ll sqr(ll x){return x*x;}const int N=55;ll n,x[N];ll a,b;ll f[N];int main(){freopen("t.in","r",stdin);freopen("t1.out","w",stdout);read(n); read(a); read(b);for (int i=1;i<=n;i++) read(x[i]);sort(x+1,x+n+1);n=unique(x+1,x+n+1)-x-1;memset(f,0x3f,sizeof(f));f[0]=0;for (int i=1;i<=n;i++)for (int j=0;j<i;j++)f[i]=min(f[i],f[j]+a+b*sqr(x[i]-x[j+1]));printf("%lld\n",f[n]);return 0;}</span>

状压dp

<span style="font-family:Microsoft YaHei;font-size:14px;color:#000066;">#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;inline char nc(){static char buf[100000],*p1=buf,*p2=buf;if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }return *p1++;}inline void read(int &x){char c=nc(),b=1;for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline ll sqr(int x){return x*x;}const int N=21;int n,x[N];int a,b;ll f[1<<N];inline ll dp(int sta){if (f[sta]<1e15) return f[sta];int minv,maxv,cur;for (int i=1;i<=n;i++){if (sta&(1<<(i-1))) continue;cur=sta; minv=1<<30; maxv=-1<<30;for (int j=i;j<=n;j++){if (sta&(1<<(j-1))) continue;cur|=1<<(j-1); maxv=max(maxv,x[j]); minv=min(minv,x[j]);f[sta]=min(f[sta],dp(cur)+a+sqr(maxv-minv)*b);}}return f[sta];}int main(){freopen("t.in","r",stdin);freopen("t2.out","w",stdout);read(n); read(a); read(b);for (int i=1;i<=n;i++) read(x[i]);memset(f,0x3f,sizeof(f));f[(1<<n)-1]=0;printf("%lld\n",dp(0));return 0;}</span>

正解

<span style="font-family:Microsoft YaHei;font-size:14px;color:#000066;">#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;inline char nc(){static char buf[100000],*p1=buf,*p2=buf;if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }return *p1++;}inline void read(int &x){char c=nc(),b=1;for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}inline ll sqr(int x){return x*x;}const int N=21;int sx[N],icnt;inline int Bin(int x){return lower_bound(sx+1,sx+icnt+1,x)-sx;}int n,x[N];int a,b;ll ans,f[N][N][N][N],g[N][N];int main(){int l,r,down,up,tl,tr,maxv,minv;freopen("t.in","r",stdin);freopen("t.out","w",stdout);read(n); read(a); read(b);for (int i=1;i<=n;i++) read(x[i]),sx[++icnt]=x[i];sort(sx+1,sx+icnt+1); icnt=unique(sx+1,sx+icnt+1)-sx-1;for (int i=1;i<=n;i++) x[i]=Bin(x[i]);for (int i=1;i<=n;i++)for (int j=1;j+i-1<=n;j++){l=j,r=j+i-1;maxv=-1<<30,minv=1<<30;for (int k=l;k<=r;k++) maxv=max(maxv,x[k]),minv=min(minv,x[k]);g[l][r]=a+sqr(sx[maxv]-sx[minv])*b;for (down=1;down<=icnt;down++)for (up=down;up<=icnt;up++){if (up<minv || down>maxv) continue;tl=l-1,tr=r+1;for (int k=l;k<=r;k++) if (down<=x[k] && x[k]<=up) tl=k; else break;for (int k=r;k>=l;k--) if (down<=x[k] && x[k]<=up) tr=k; else break;if (tl>=tr)f[l][r][down][up]=0;elsef[l][r][down][up]=g[tl+1][tr-1];for (int k=l;k<r;k++)f[l][r][down][up]=min(f[l][r][down][up],f[l][k][down][up]+f[k+1][r][down][up]);g[l][r]=min(g[l][r],f[l][r][down][up]+a+sqr(sx[up]-sx[down])*b);}for (down=maxv+1;down<=icnt;down++)for (up=down;up<=icnt;up++)f[l][r][down][up]=g[l][r];for (up=1;up<minv;up++)for (down=1;down<=up;down++)f[l][r][down][up]=g[l][r];}printf("%lld\n",g[1][n]);return 0;}</span>

T2

Trie套vector 自己都不知道对不对

T3

前两个点就是环  第三个点线段很短 可以当做点 其他么 贪心啊 模拟退火啊 随机排列啊 取个较优的吧 根本没想过把线段割开