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https://blog.csdn.net/qq574857122/article/details/34430283
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Meaning of the questions:
Given a sequence of length n
To give a sequence of fixed length k
LCIS and outputs the required sequence
If multiple solutions Solution output random. .
= - = knock when listening to the pre meaning little song not want to know. Never occurred before. .
。
#include<stdio.h> #include<iostream> #include<string.h> #include<set> #include<vector> #include<map> #include<math.h> #include<string> #include<stdlib.h> #include<algorithm> using namespace std; #define N 505 int a[N],s[N]; int n, k; int dp[N][N]; int pre[N][N][2]; int is[N][N]; vector<int>G; int main(){ int i,j,co; while(~scanf("%d",&n)){ G.clear(); for(i=1;i<=n;i++)scanf("%d",&a[i]); scanf("%d",&k); for(i=1;i<=k;i++)scanf("%d",&s[i]); memset(dp, 0, sizeof dp); memset(pre, 0, sizeof pre); memset(is, 0, sizeof is); for(i=1;i<=n;i++) { int maxx = 0; int x = 0, y = 0; for(j=1;j<=k;j++) { dp[i][j] = dp[i-1][j]; if(a[i]>s[j] && maxx < dp[i-1][j]) { maxx = dp[i-1][j]; if(is[i-1][j]) x = i-1, y = j; else { x = pre[i-1][j][0]; y = pre[i-1][j][1]; } } if(a[i]==s[j]) { is[i][j] = 1; dp[i][j] = maxx + 1; if(is[x][y]) pre[i][j][0] = x, pre[i][j][1] = y; else { pre[i][j][0] = pre[x][y][0]; pre[i][j][1] = pre[x][y][1]; } continue; } if(is[i-1][j]) pre[i][j][0] = i-1, pre[i][j][1] = j; else { pre[i][j][0] = pre[i-1][j][0]; pre[i][j][1] = pre[i-1][j][1]; } } } int ans = 0; int x = n, y = k; for(i=1;i<=k;i++)if(ans<dp[n][i]){ ans = dp[n][i]; x = n, y = i; } printf("%d\n",ans); if(!ans)continue; while(x+y) { if(is[x][y])G.push_back(a[x]); int x1 = pre[x][y][0]; int y1 = pre[x][y][1]; x = x1 , y = y1; } for(i=G.size()-1; i>=0; i--){ printf("%d",G[i]); i?
printf(" "):puts(""); } } return 0; } /* 2 1 2 3 1 2 3 */