[AcWing 1015] 摘花生

BFS + 动态规划 运行时间 855 ms

---

点击查看代码

#include<iostream>
#include<cstring>
#include<queue>

#define fi first
#define se second

using namespace std;

typedef pair<int,int> PII;

const int N = 110;

int t, r, c;
int f[N][N], g[N][N];
int dx[] = {1, 0}, dy[] = {0, 1};
bool st[N][N];

void solve()
{
	memset(f, 0, sizeof f);
	for (int i = 1; i <= r; i ++)
		for (int j = 1; j <= c; j ++)
			st[i][j] = false;
	queue<PII> q;
	q.push({1, 1});
	st[1][1] = true;
	f[1][1] = g[1][1];
	while (q.size()) {
		auto u = q.front();
		q.pop();
		for (int i = 0; i < 2; i ++) {
			int x = u.fi + dx[i], y = u.se + dy[i];
			if (x <= r && y <= c) {
				f[x][y] = max(f[x][y], f[u.fi][u.se] + g[x][y]);
				if (!st[x][y]) {
					q.push({x, y});
					st[x][y] = true;
				}
			}
		}
	}
	cout << f[r][c] << endl; 
}

int main()
{
	cin >> t;
	while (t --) {
		cin >> r >> c;
		for (int i = 1; i <= r; i ++)
			for (int j = 1; j <= c; j ++)
				scanf("%d", &g[i][j]);
		solve();
	}
	return 0;
}

---

动态规划 运行时间 382 ms

---

点击查看代码

#include<iostream>
#include<cstring>

using namespace std;

const int N = 110;

int t;
int g[N][N], f[N][N];

int main()
{
	cin >> t;
	while (t --) {
		int r, c;
		cin >> r >> c;
		for (int i = 1; i <= r; i ++)
			for (int j = 1; j <= c; j ++)
				scanf("%d", &g[i][j]);
		memset(f, 0, sizeof f);
		for (int i = 1; i <= r; i ++)
			for (int j = 1; j <= c; j ++)
				f[i][j] = max(f[i - 1][j], f[i][j - 1]) + g[i][j];
		cout << f[r][c] << endl;
	}	
	return 0;
}

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1. 状态表示
   $f[i][j]$ 表示从 $(1,1)$ 走到 $(i,j)$ 的所有路线的最大值
2. 状态计算
   $f[i][j] = max(f[i - 1][j], f[i][j - 1]) + g[i][j]$