Uvalive 4725 airport [DP]

Question connection

The DP status of this question is DP [I] [J], which indicates that the first apron of the previous I time has the maximum value when J planes fly out. Because there are only two la s, the first one is only OK, and the second one is OK.

Of course, there is still a situation where the plane is not flying enough in this question. This can be handled in advance (0, 0 and all the planes in front are flying)

Think about the trouble, but it is not unsolvable.

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define N 5010int n;struct note{    int a,b;}dat[N];int sum[2][N];int dp[N][N];int min(int a,int b){    if(a == -1) return b;    if(a < b) return a;return b;}int main(){    int t;    cin >> t;    while(t--)    {        cin >> n;        for(int i = 1;i <= n;i++) scanf("%d%d",&dat[i].a,&dat[i].b);        for(int i = 0;i <= n;i++) for(int j = 0;j <= n;j++) dp[i][j] = -1;        dp[0][0] = 0;        sum[0][0] = sum[1][0] = 0;        for(int i = 1;i <= n;i++)        {            sum[0][i] = sum[0][i-1] + dat[i].a;            sum[1][i] = sum[1][i-1] + dat[i].b;        }        int kk = 0;        for(int i = 1;i <= n;i++)        {            if(dat[i].a == 0 && dat[i].b == 0 && sum[0][i] + sum[1][i] <= kk)            {                continue;            } else            {                kk++;                dat[kk].a = dat[i].a;                dat[kk].b = dat[i].b;            }        }        n = kk;        for(int i = 1;i <= n;i++)        {            sum[0][i] = sum[0][i-1] + dat[i].a;            sum[1][i] = sum[1][i-1] + dat[i].b;        }        //cout << n << " ";        for(int i = 0;i < n;i++)        {            for(int j = 0;j <= i;j++)            {                if(dp[i][j] == -1) continue;                if(sum[0][i+1] >= j+1)                   dp[i+1][j+1] = min(     dp[i+1][j+1],                         max(dp[i][j],max(sum[0][i+1] - j,sum[1][i+1] - (i - j))));                if(sum[1][i+1] >= i-j+1)                   dp[i+1][j] = min(    dp[i+1][j],                         max(dp[i][j],max( sum[0][i+1] - j,sum[1][i+1] - (i - j) )));            }        }        int mi = 99999999;        for(int i = 0;i <= n;i++) if(dp[n][i] != -1) mi = min(mi,dp[n][i]);        if(mi < 1) mi = 1;        printf("%d\n",mi-1);    }    return 0;}