hdu 4374 单调队列优化动态规划

思路：我只想说，while(head<=rear&&que[rear].val+sum[j]-sum[que[rear].pos-1]<=dp[i-1][j]+num[i-1][j])表达式中，我把最后的num[i-1][j]丢了，检查了一整天啊！一整天！我那丢失的时间！

寻找从上一层到该层中最优的转换点，如果该转化点到该点超过t就出队。

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define Maxn 110
#define Maxm 10010
#define inf 1000000000
using namespace std;
int dp[Maxn][Maxm],sum[Maxm],num[Maxn][Maxm];
int que[Maxm*5];
void init()
{
    memset(sum,0,sizeof(sum));
    memset(num,0,sizeof(num));
}
inline int ReadInt()
{
    int flag = 1;
    char ch;
    int a = 0;
    while((ch = getchar()) == ' ' || ch == '\n');
    if(ch == '-') flag = -1;
    else
    a += ch - '0';
    while((ch = getchar()) != ' ' && ch != '\n')
    {
        a *= 10;
        a += ch - '0';
    }
    return flag * a;
}
int main()
{
    int n,m,x,t,i,j,head,rear;
    while(scanf("%d%d%d%d",&n,&m,&x,&t)!=EOF){
        init();
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                num[i][j]=ReadInt();
                dp[i][j]=-inf;
            }
        }
        memset(dp[0],-50,sizeof(dp));
        dp[0][x]=0;
        for(i=1;i<=n;i++){
            memset(sum,0,sizeof(sum));
            head=1,rear=0;
            for(j=1;j<=m;j++){
                sum[j]=sum[j-1]+num[i][j];
                while(head<=rear&&dp[i-1][que[rear]]+sum[j]-sum[que[rear]-1]<=dp[i-1][j]+num[i][j])
                    rear--;
                que[++rear]=j;
                while(j-que[head]>t&&head<=rear)
                    head++;
                if(head<=rear)
                dp[i][j]=dp[i-1][que[head]]+sum[j]-sum[que[head]-1];
            }
            head=1,rear=0;
            memset(sum,0,sizeof(sum));
            for(j=m;j>=1;j--){
                sum[j]=sum[j+1]+num[i][j];
                while(head<=rear&&dp[i-1][que[rear]]+sum[j]-sum[que[rear]+1]<=dp[i-1][j]+num[i][j])
                    rear--;
                que[++rear]=j;
                while(que[head]-j>t&&head<=rear)
                    head++;
                if(head<=rear)
                dp[i][j]=max(dp[i][j],dp[i-1][que[head]]+sum[j]-sum[que[head]+1]);
            }
        }
        int ans=-inf;
        for(i=1;i<=m;i++)
            ans=max(ans,dp[n][i]);
        printf("%d\n",ans);
    }
    return 0;
}