牛客国庆集训派对Day6 - E DP

版权声明：欢迎随便转载。 https://blog.csdn.net/a1214034447/article/details/82988673

题目链接:点击这里

解题思路:

假设现在处于(x,y)位置刚获得一个奖励,那么肯定x∈(a数组),y∈(b数组),所以将a数组,b数组,离散化变成矩阵,那么

dp[i][j] = max(dp[i-1][j],r,dp[i][j-1]+c),因为要到达(i,j)第一种先到达i,第二种是先到达j.

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int mx = 1e3 + 10;
typedef pair <int,int> pa;
map <pair<int,int>,ll> mp;
int xi[mx],yi[mx],n,m;
ll r[mx][mx],c[mx][mx],dp[mx][mx];
int main()
{      
    scanf("%d%d",&n,&m);
    int a,b,z;
    for(int i=1;i<=n;i++){
        scanf("%d%d%d",&a,&b,&z);
        if(mp.count(pa(a,b))) mp[pa(a,b)] += z;
        else mp[pa(a,b)] = z;
        xi[i] = a,yi[i] = b;
    }
    sort(xi+1,xi+1+n);
    sort(yi+1,yi+1+n);
    int lonx = unique(xi+1,xi+1+n) - xi;
    int lony = unique(yi+1,yi+1+n) - yi;
    for(int i=1;i<lonx;i++){
        for(int j=1;j<lony;j++){
            if(mp.count(pa(xi[i],yi[j])))
            r[i][j] = r[i][j-1] + mp[pa(xi[i],yi[j])];
            else r[i][j] = r[i][j-1];
        }
    }
    for(int j=1;j<lony;j++){
        for(int i=1;i<lonx;i++){
            if(mp.count(pa(xi[i],yi[j])))
            c[i][j] = c[i-1][j] + mp[pa(xi[i],yi[j])];
            else c[i][j] = c[i-1][j];
        }
    }
    for(int i=1;i<lonx;i++){
        for(int j=1;j<lony;j++){
            if(xi[i]+yi[j]<=m){
                dp[i][j] = max(dp[i][j],dp[i-1][j]+r[i][j]*(m-xi[i]-yi[j]+1));
                dp[i][j] = max(dp[i][j],dp[i][j-1]+c[i][j]*(m-xi[i]-yi[j]+1));
            }
        }
    }
    ll ans = 0;
    for(int i=1;i<lonx;i++){
        for(int j=1;j<lony;j++){
            ans = max(ans,dp[i][j]);
        }
    }
    printf("%lld\n",ans);
    return 0;
}