P1091 合唱隊(duì)形

看一下題解吧，你好i需要正反搜一下lcs ,然后合并

#include<bits/stdc++.h> using namespace std; #define LOACL freopen("in","r",stdin);\freopen("out","w",stdout); const int inf = 987654321; const int sz = 1e6 + 5; const int mod = 1e9 + 7; const int sqrtn = 300; #define add(u,v,w) (e[++tot]=(edge){v,head[u],1},head[u]=tot;) #define f(i,l,r) for(int i=(int)l;i<=(int)r;++i) #define g(i,l,r) for(int i=(int)l;i>=(int)r;--i) #define CLR(arr,val) memset(arr,val,sizeof(arr)) typedef long long ll; int n,a[sz],dp[2][sz],ans; int main() {LOACLcin>>n;f(i,1,n)cin>>a[i];f(i,1,n)f(j,0,i-1)if(a[i]>a[j])dp[0][i]=max(dp[0][i],dp[0][j]+1); g(i,n,1)g(j,n+1,i+1) if(a[i]>a[j]) dp[1][i]=max(dp[1][i],dp[1][j]+1);f(i,1,n) ans=max(ans,dp[0][i]+dp[1][i]-1);cout<<n-ans<<endl;return 0; } View Code

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P1280 尼克的任務(wù)

倒著搜索尋找是否能休息?

轉(zhuǎn)移方程 : f[i]=f[i+1]+1;

　　　　　d[i]=f[i+s];? i 時(shí)間內(nèi)有任務(wù) 需要向后找狀態(tài)

#include<bits/stdc++.h> using namespace std; #define LOACL freopen("in","r",stdin);\freopen("out","w",stdout); const int inf = 987654321; const int sz = 1e6 + 5; const int mod = 1e9 + 7; const int sqrtn = 300; #define add(u,v,w) (e[++tot]=(edge){v,head[u],1},head[u]=tot;) #define f(i,l,r) for(int i=(int)l;i<=(int)r;++i) #define g(i,l,r) for(int i=(int)l;i>=(int)r;--i) #define CLR(arr,val) memset(arr,val,sizeof(arr)) typedef long long ll; ll n,m,num=1; ll sum[sz],f[sz]; struct node {ll l,r; }e[sz]; bool cmp(node l,node r) {return l.l>r.l; } int main() {LOACLcin>>n>>m;f(i,1,m){cin>>e[i].l>>e[i].r ;sum[e[i].l]++;}sort(e+1,e+m+1,cmp);g(i,n,1){if(sum[i]==0)f[i]=f[i+1]+1;else {f(j,1,sum[i]){if(f[i]<f[i+e[num].r])f[i]=f[i+e[num].r]; num++;}} }cout<<f[1]<<endl; return 0; } View Code

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P1880 [NOI1995]石子合并

處理一下 前綴和 因?yàn)槭且粋€(gè)環(huán) 所以2倍

　狀態(tài)轉(zhuǎn)移 dp[i][j]=max/min(dp[i][k]+dp[k+1][j]+sum(i,j));

#include<bits/stdc++.h>using namespace std;#define LOACL freopen("in","r",stdin);\freopen("out","w",stdout); const int inf = 987654321;const int sz = 1e6 + 5;const int mod = 1e9 + 7;const int sqrtn = 600; #define add(u,v,w) (e[++tot]=(edge){v,head[u],1},head[u]=tot;) #define f(i,l,r) for(int i=(int)l;i<=(int)r;++i)#define g(i,l,r) for(int i=(int)l;i>=(int)r;--i)#define CLR(arr,val) memset(arr,val,sizeof(arr)) typedef long long ll; int n,a[300],f1[300][300],f2[300][300],sum[300];int main(){LOACLcin>>n;f(i,1,n){cin>>a[i];a[i+n]=a[i];}f(i,1,n<<1){sum[i]=sum[i-1]+a[i];}g(i,n<<1,1){f(j,i+1,min(n+n,i+n-1)){f2[i][j]=inf;f(k,i,j-1){f1[i][j]=max(f1[i][j],f1[i][k]+f1[k+1][j]+sum[j]-sum[i-1]);f2[i][j]=min(f2[i][j],f2[i][k]+f2[k+1][j]+sum[j]-sum[i-1]);}}}// f2 min int maxn = -inf,minn = inf;f(i,1,n){maxn =max(maxn,f1[i][i+n-1]);minn =min(minn,f2[i][i+n-1]); }cout <<minn<<endl;cout <<maxn<<endl;return 0;} View Code

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轉(zhuǎn)載于:https://www.cnblogs.com/corx/p/8580292.html

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