ACM模版
描述
题解
image.png代码
代码语言：javascript复制#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX_LEN_NAME = 11;
const int MAXN = 1010;
struct Node
{
int x, y;
bool operator < (const Node &rhs) const
{
if (x != rhs.x)
{
return x > rhs.x;
}
return y < rhs.y;
}
} arr[MAXN];
int n;
int dp[MAXN][MAXN >> 1];
int cost[MAXN][MAXN >> 1];
char name[MAX_LEN_NAME];
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
scanf("%d%s", &n, name);
int sum = 0;
for (int i = 1; i <= n; i)
{
scanf("%d%d", &arr[i].x, &arr[i].y);
sum = arr[i].x;
}
sort(arr 1, arr 1 n);
memset(dp, 0, sizeof(dp));
memset(cost, 0, sizeof(cost));
int cur = 0;
int i = name[0] == 'P' ? 2 : 1;
for (; i <= n; i)
{
cur;
int t = (cur 1) >> 1;
for (int j = 1; j <= t; j)
{
int &ans = dp[i][j] = dp[i - 1][j];
cost[i][j] = cost[i - 1][j];
if (j == 1 || dp[i - 1][j - 1])
{
int tmp = dp[i - 1][j - 1] arr[i].y;
if (tmp > ans)
{
ans = tmp;
cost[i][j] = cost[i - 1][j - 1] arr[i].x;
}
else if (tmp == ans)
{
cost[i][j] = min(cost[i][j], cost[i - 1][j - 1] arr[i].x);
}
}
}
}
printf("%d %dn", sum - cost[n][(cur 1) >> 1], dp[n][(cur 1) >> 1]);
}
return 0;
}
