题意:n*n的方格,从左上角到右下角两次。每一次经过的路径中,如果有数字,数字都会变成0并计数。求两次路径的最大计数。
思路:线性dp,从左上角到右下角步数固定为 2 * n - 2步。 初始时0步dp[0][1][1] = grid[1][1],知道了x1和x2可以确定对应的y,可以直接进行状态转移。 可以增加剪枝:x <= min(n, s1 + 1)。
总结:
1 忽略了当到了终点时x1 == x2的情况,被直接跳过了。
2 忽略了x不动,当前步数都是由y移动的情况(4种转移状态之一)
void solve(){
int n;
cin >> n;
vector<vector<int>> grid(n + 1, vector<int>(n + 1, 0));
{
int x, y, k;
while (cin >> x >> y >> k && (x && y && k)){
grid[x][y] = k;
}
}
vector<vector<vector<int>>> dp(2, vector<vector<int>>(n + 1, vector<int>(n + 1)));
dp[0][1][1] = grid[1][1];
int p = 0;
for (int s = 1; s <= 2 * n - 2; ++s){
p ^= 1;
for (int x1 = 1; x1 <= min(n, s + 1); ++x1){
for (int x2 = 1; x2 <= min(n, s + 1); ++x2){
if ((x1 != x2) || s == 2 * n - 2){
int y1 = s + 2 - x1;
int y2 = s + 2 - x2;
if (y1 <= n && y2 <= n){
auto& cur = dp[p][x1][x2];
if (x1 > 1 && x2 > 1){
cur = max(cur, dp[p ^ 1][x1 - 1][x2 - 1] + grid[x1][y1] + grid[x2][y2]);
}
if (x1 > 1 && y2 > 1){
cur = max(cur, dp[p ^ 1][x1 - 1][x2] + grid[x1][y1] + grid[x2][y2]);
}
if (x2 > 1 && y1 > 1){
cur = max(cur, dp[p ^ 1][x1][x2 - 1] + grid[x1][y1] + grid[x2][y2]);
}
if (y1 > 1 && y2 > 1){
cur = max(cur, dp[p ^ 1][x1][x2] + grid[x1][y1] + grid[x2][y2]);
}
}
}
}
}
}/*
for (int i = 1; i <= n; ++i){
for (int j = 1; j <= n; ++j){
cout << grid[i][j] << " \n"[j == n];
}
}
for (int i = 1; i <= n; ++i){
for (int j = 1; j <= n; ++j){
cout << dp[p][i][j] << " \n"[j == n];
}
}*/
cout << dp[p][n][n] - grid[n][n] << '\n';
}
标签:NOIP2000,cur,P1004,取数,grid,&&,x2,x1,dp
From: https://www.cnblogs.com/yxcblogs/p/18114114