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蓝桥杯--算法训练 方格取数(多线程dp)

2016-03-13 15:22 357 查看

题意

取过的数会变成0,因此两条路径相交的部分只能取一次。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int maxn = 12;
int   dp[maxn][maxn][maxn][maxn], g[maxn][maxn];
int   n;

void dpSolve()
{
/*for(int k = 2; k < n + n; ++k)
{
for(int x1 = 1; x1 <= n; ++x1){
for(int x2 = 1; x2 <= n; ++x2){

int y1 = k - x1, y2 = k - x2;
if(y1 < 1 || y1 > n || y2 < 1 || y2 > n) continue;
dp[k][x1][x2] = max(max(dp[k - 1][x1][x2], dp[k - 1][x1 - 1][x2 - 1]),
max(dp[k - 1][x1 - 1][x2], dp[k - 1][x1][x2 - 1]));
if(y1 == y2) dp[k][x1][x2] += g[x1][y1];
else dp[k][x1][x2] += g[x1][y1] + g[x2][y2];
}
}
}
cout << dp[n + n - 1]
[n - 1] << endl;*/
for(int x1 = 1; x1 <= n; ++x1)
{
for(int y1 = 1; y1 <= n; ++y1)
for(int x2 = 1; x2 <= n; ++x2)
for(int y2 = 1; y2 <= n; ++y2){
dp[x1][y1][x2][y2] = max(max(dp[x1-1][y1][x2-1][y2], dp[x1-1][y1][x2][y2-1]),
max(dp[x1][y1-1][x2-1][y2],dp[x1][y1-1][x2][y2-1]));
if(x1 == x2 && y1 == y2) dp[x1][y1][x2][y2] += g[x1][y2];
else dp[x1][y1][x2][y2] += g[x1][y1] + g[x2][y2];
}
}
cout << dp

<< endl;
}

int main()
{
#ifdef LOCAL
freopen("data.in", "r", stdin);
#endif // LOCAL
cin >> n;
int a, b, c;
while(cin >> a >> b >> c)
{
if(!a && !b && !c) break;
g[a][b] = c;
}
dpSolve();
return 0;
}
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标签:  蓝桥杯 dp middle