9.15 DP合集水表
显然难了一些啊。
凸多边形的三角剖分
瞄了一眼题解。
和蛤蛤的烦恼一样，裸的区间dp。
设f[i][j]表示i~j的点三角剖分最小代价。
显然(f[i][i+1]=0,f[i][i+2]=w[i]*w[i+1]*w[i+2])
然后枚举i,j和哪个点剖。
[f[l][r]=min(f[l][r],f[l][i]+f[i][r]+w[l]*w[r]*w[i]) (l<i<r)
]
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "division"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
rg int x=0;rg char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
ll w[101],f[101][101];
int main(){
#ifdef xzz
freopen(Fname".in","r",stdin);
freopen(Fname".out","w",stdout);
#endif
int n=gi(),r;
rep(i,1,n)w[i]=w[i+n]=gi();
n*=2;
rep(i,3,n)f[i-2][i]=w[i-2]*w[i-1]*w[i];
rep(siz,4,n/2)drep(l,n-siz+1,1){
r=l+siz-1;
f[l][r]=1e18;
rep(i,l+1,r-1)f[l][r]=min(f[l][r],f[l][i]+f[i][r]+w[l]*w[r]*w[i]);
}
n/=2;
ll ans=2e18;
rep(i,1,n)ans=min(ans,f[i][i+n-1]);
printf("%lld
",ans);
return 0;
}
求最长公共子序列
字符序列的子序列是指从给定字符序列中随意地（不一定连续）去掉若干个字符（可能一个也不去掉）后所形成的字符序列。
SBT，切掉
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#include<cstring>
#define Fname "lcs"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
rg int x=0;rg char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
char a[5002],b[5002];
int f[5002][5002];
int main(){
#ifdef xzz
freopen(Fname".in","r",stdin);
freopen(Fname".out","w",stdout);
#endif
scanf("%s%s",a+1,b+1);
int lena=strlen(a+1)-1,lenb=strlen(b+1)-1;
rep(i,1,lena)rep(j,1,lenb)
if(a[i]==b[j])f[i][j]=f[i-1][j-1]+1;
else f[i][j]=max(f[i-1][j],f[i][j-1]);
printf("%d
",f[lena][lenb]);
return 0;
}
卡车更新问题
设f[i][j]表示第i天，上次更新是j就好了
主要难在输出方案，恶心细节多
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "truck"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
rg int x=0;rg char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
#define db double
db R[23],U[23],C[23];
db f[23][23];
int g[23][23];
il vd Min(db&a,int&b,db c,int d){if(c>a)a=c,b=d;}
//f[i][j] 到第i天，上次更新是第j天的总回收额max
bool yes[23];
int main(){
#ifdef xzz
freopen(Fname".in","r",stdin);
freopen(Fname".out","w",stdout);
#endif
int n=gi(),k=gi();
rep(i,0,k)scanf("%lf",&R[i]);
rep(i,0,k)scanf("%lf",&U[i]);
rep(i,0,k)scanf("%lf",&C[i]);
rep(i,1,n+1)rep(j,0,k)f[i][j]=-233333.33;
f[1][0]=R[0]-U[0];
rep(i,1,n-1)rep(j,0,i-1){
if(i-j<k)Min(f[i+1][j],g[i+1][j],f[i][j]+R[i-j]-U[i-j],j);
Min(f[i+1][i],g[i+1][i],f[i][j]+R[0]-U[0]-C[i-j],j);
}
int ans=n-1;
rep(i,n-k,n-2)if(f[n][i]>f[n][ans])ans=i;
printf("%.1f
",f[n][ans]);
int x=n,y=ans;
while(x){
if(y==x-1)yes[x]=1;
y=g[x][y],--x;
}x=n,y=ans;
printf("%d 0 %.1f
",1,R[0]-U[0]);
y=1;
rep(i,2,n){
if(yes[i])printf("%d 1 %.1f
",i,R[0]-U[0]-C[y]),y=1;
else printf("%d 0 %.1f
",i,R[y]-U[y]),++y;
}
return 0;
}
选课
裸的树形背包但要输出方案。GG
不会
为了挽尊，特写spj一个。。。
#include<bits/stdc++.h>
using namespace std;
int main(int argc,char *argv[])
{
FILE *in,*out,*ans,*score,*report;
out=fopen(argv[2],"r");
ans=fopen(argv[3],"r");
score=fopen(argv[5],"w");
report=fopen(argv[6],"w");
int a=-2333,b=-6666;
fscanf(out,"%d",&a);
fscanf(ans,"%d",&b);
if(a==b){
while(fscanf(out,"%d",&a)==1||fscanf(ans,"%d",&b)==1)
if(a!=b){fprintf(score,"3"),fprintf(report,"方案错了。。。3分");return 0;}
fprintf(score,"10"),fprintf(report,"泥太有才了！");
}
else fprintf(score,"0"),fprintf(report,"答案都错了。。。期望%d但输出%d",b,a);
}
没方案或方案错了就会得到3分233
贴一下没方案的代码
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "course"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
rg int x=0;rg char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
const int maxn=520,maxm=maxn;
int m,n,w[maxn],fir[maxn],dis[maxm],nxt[maxm],id;
il vd add(int a,int b){nxt[++id]=fir[a],fir[a]=id,dis[id]=b;}
int f[maxn][maxn];
il vd dp(int x,int M){
if(M==0)return;
erep(i,x){
rep(j,1,M-1)f[dis[i]][j]=f[x][j];
dp(dis[i],M-1);
drep(j,M,1)f[x][j]=max(f[x][j],f[dis[i]][j-1]+w[dis[i]]);
}
}
int main(){
freopen(Fname".in","r",stdin);
freopen(Fname".out","w",stdout);
m=gi(),n=gi();
rep(i,1,m)add(gi(),i),w[i]=gi();
dp(0,n);
printf("%d
",f[0][n]);
return 0;
}
蛤蛤的烦恼
利用了凸多边形的性质。
然而窝不会证。
f[l][r]表示从l开始遍历l~r的最小代价。
g[l][r]表示从r开始遍历l~r的最小代价。
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#include<cmath>
#define Fname "frog"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
rg int x=0;rg char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
#define db double
#define maxn 721
db x[maxn],y[maxn];
db f[maxn][maxn],g[maxn][maxn];
il db dist(int a,int b){return sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]));}
int main(){
#ifdef xzz
freopen(Fname".in","r",stdin);
freopen(Fname".out","w",stdout);
#endif
int n=gi(),r;
rep(i,1,n)scanf("%lf%lf",&x[i],&y[i]);
rep(siz,2,n)drep(l,n-siz+1,1){
r=l+siz-1;
f[l][r]=min(f[l+1][r]+dist(l,l+1),g[l+1][r]+dist(l,r));
g[l][r]=min(g[l][r-1]+dist(r-1,r),f[l][r-1]+dist(l,r));
}
printf("%.3f
",f[1][n]);
return 0;
}
警卫安排
灰常经典乐，以前早写过类似的辣
f[i]表示父亲染i，g表示i染i，h表示i的儿子染i
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "security"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
rg int x=0;rg char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
const int maxn=722;
int n,w[maxn];
int fir[maxn],dis[maxn],nxt[maxn],id;
il vd add(int a,int b){nxt[++id]=fir[a],fir[a]=id,dis[id]=b;}
int f[maxn],g[maxn],h[maxn];
//f父亲or自己染自己，g自己染自己，h儿子or自己染自己
il vd dp(int x){
f[x]=0,g[x]=w[x],h[x]=1e9;
int Sum=0;
erep(i,x){
dp(dis[i]);
Sum+=h[dis[i]];
g[x]+=f[dis[i]];
f[x]+=h[dis[i]];
}
erep(i,x)h[x]=min(h[x],Sum-h[dis[i]]+g[dis[i]]);
h[x]=min(h[x],g[x]);
f[x]=min(f[x],g[x]);
}
int isson[maxn];
int main(){
#ifdef xzz
freopen(Fname".in","r",stdin);
freopen(Fname".out","w",stdout);
#endif
n=gi();
int son,s,a,root;
rep(i,1,n){
s=gi();
w[s]=gi();
son=gi();
while(son--)a=gi(),isson[a]=1,add(s,a);
}
rep(i,1,n)if(!isson[i])root=i;
dp(root);
printf("%d
",min(g[root],h[root]));
return 0;
}
最长上升子序列
SBT。k前去掉比a[k]打的，k后去掉比a[k]小的，显然答案不变而且能取到k
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "lis"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
rg int x=0;rg char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
const int maxn=2e5+2;
int d[maxn],a[maxn];
int main(){
#ifdef xzz
freopen(Fname".in","r",stdin);
freopen(Fname".out","w",stdout);
#endif
int s,n=gi(),k=gi(),maxf=0;
rep(i,1,n)a[i]=gi();
d[0]=-2e9;
rep(i,1,n)d[i]=2e9;
rep(i,1,n){
if((i<k&&a[i]>=a[k])||(i>k&&a[i]<=a[k]))continue;
s=lower_bound(d,d+maxf+1,a[i])-d;
//f[i]=s
d[s]=min(d[s],a[i]);maxf=max(maxf,s);
}
printf("%d
",maxf);
return 0;
}
最短回文串
同编辑距离。稍作改动就行
f[l][r]表示[l,r]变成回文串最少插几个字符
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define Fname "palindrome"
using namespace std;
#define rep(a,b,c) for(rg int a=b;a<=c;a++)
#define drep(a,b,c) for(rg int a=b;a>=c;a--)
#define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
#define il inline
#define rg register
#define vd void
typedef long long ll;
il int gi(){
rg int x=0;rg char ch=getchar();
while(ch<'0'||ch>'9')ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x;
}
char s[5010];
short f[5010][5010];
int main(){
#ifdef xzz
freopen(Fname".in","r",stdin);
freopen(Fname".out","w",stdout);
#endif
int n=gi(),r;
scanf("%s",s+1);
rep(siz,2,n)drep(l,n-siz+1,1){
r=l+siz-1;
if(s[l]==s[r])f[l][r]=f[l+1][r-1];
else f[l][r]=min(f[l+1][r],f[l][r-1])+1;
}printf("%d
",f[1][n]);
return 0;
}
水表搞定了！
备份，密码同博客密码