题目连接：https://codeforces.com/contest/1517/problem/C
Fillomino is a classic logic puzzle. (You do not need to know Fillomino in order to solve this problem.) In one classroom in Yunqi town, some volunteers are playing a board game variant of it:
Consider an n by n chessboard. Its rows are numbered from 1 to n from the top to the bottom. Its columns are numbered from 1 to n from the left to the right. A cell on an intersection of x-th row and y-th column is denoted (x,y). The main diagonal of the chessboard is cells (x,x) for all 1≤x≤n.
A permutation of {1,2,3,…,n} is written on the main diagonal of the chessboard. There is exactly one number written on each of the cells. The problem is to partition the cells under and on the main diagonal (there are exactly 1+2+…+n such cells) into n connected regions satisfying the following constraints:
Every region should be connected. That means that we can move from any cell of a region to any other cell of the same region visiting only cells of the same region and moving from a cell to an adjacent cell.
The x-th region should contain cell on the main diagonal with number x for all 1≤x≤n.
The number of cells that belong to the x-th region should be equal to x for all 1≤x≤n.
Each cell under and on the main diagonal should belong to exactly one region.
Input
The first line contains a single integer n (1≤n≤500) denoting the size of the chessboard.
The second line contains n integers p1, p2, …, pn. pi is the number written on cell (i,i). It is guaranteed that each integer from {1,…,n} appears exactly once in p1, …, pn.
Output
If no solution exists, output ?1.
Otherwise, output n lines. The i-th line should contain i numbers. The j-th number on the i-th line should be x if cell (i,j) belongs to the the region with x cells.
Examples
input
3
2 3 1
output
2
2 3
3 3 1
input
5
1 2 3 4 5
output
1
2 2
3 3 3
4 4 4 4
5 5 5 5 5
分析
对角线每个位置每个位置能向右延伸就向右，不能就向下延伸，必定可以实现。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;int n;
int vis[505][505],d[505],a[505][505];
int dirx[2]={
0, 1};
int diry[2]={
-1, 0};void dfs(int x,int y,int s,int step)
{
a[x][y] = s;vis[x][y] = 1;if(s == step) return;for(int i=0;i<2;i++){
int xx = x + dirx[i];int yy = y + diry[i];if(vis[xx][yy] == 0 && xx <= n && yy >0){
dfs(xx, yy, s, step + 1);break;}}
}int main() {
scanf("%d",&n);for(int i=1;i<=n;i++){
int x;scanf("%d",&x);dfs(i, i, x, 1);}for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++)printf("%d ",a[i][j]);printf("\n");}return 0;
}