洛谷P2196 && caioj 1415 动态规划6：挖地雷

没看出来动规怎么做，看到n <= 20,直接一波暴搜，过了。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;

const int MAXN = 25;
int g[MAXN][MAXN], a[MAXN], f[MAXN];
int vis[MAXN], path[MAXN], n;

int dfs(int cur)
{
	vis[cur] = 1;
	int t = 0;
	REP(i, 0, n)	
		if(!vis[i] && g[cur][i])
		{
			int temp = dfs(i);
			if(t < temp)
			{
				t = temp;
				path[cur] = i;
			}
		}
	vis[cur] = 0;
	return a[cur] + t;
}

int main()
{
	scanf("%d", &n);
	REP(i, 0, n) scanf("%d", &a[i]);
	REP(i, 0, n)
		REP(j, i + 1, n)
			scanf("%d", &g[i][j]);
	
	int ans = 0, st;
	REP(i, 0, n)
	{
		memset(path, -1, sizeof(path));
		memset(vis, 0, sizeof(vis));
		int t = dfs(i);
		if(ans < t)
		{
			st = i;
			ans = t;
			memcpy(f, path, sizeof(f));
		}
	}
	
	for(int p = st; p != -1; p = f[p]) printf("%d ", p + 1);
	puts("");
	printf("%d\n", ans);
    
    return 0;
}

然后我还是想想动规怎么做吧

因为只能往下挖，所以最后一个地窖是挖不了的
所以我们倒着推
我们设f[i]为从i开始能挖到的最大地雷
那么有
f[i] = f[j] + a[i]; i < j, i与j连接
 

#include<cstdio>
#include<cstring>
#include<algorithm>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;

const int MAXN = 25;
int g[MAXN][MAXN], a[MAXN], f[MAXN];
int path[MAXN], n;

int main()
{
	memset(path, -1, sizeof(path));
	scanf("%d", &n);
	REP(i, 0, n) scanf("%d", &a[i]);
	REP(i, 0, n)
		REP(j, i + 1, n)
			scanf("%d", &g[i][j]);
	
	int ans = 0, st;
	for(int i = n - 1; i >= 0; i--)
	{
		f[i] = a[i];
		REP(j, i + 1, n)
			if(g[i][j] && f[i] < f[j] + a[i])
			{
				f[i] = f[j] + a[i];
				path[i] = j;
			}
		if(ans < f[i])
		{
			ans = f[i];
			st = i;
		}
	}
	
	for(int p = st; p != -1; p = path[p]) printf("%d ", p + 1);
	puts("");
	printf("%d\n", ans);
    
    return 0;
}