LA 4327 Parade （DP单调队列）

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题目大意：有n+1条横线，m+1条竖线，你的任务是从最南边的路走到最北边的路，使得走过的路上的高兴值和最大。同一段路不能走超过两次，且不能从北往南走，另外每条横向路上所花的时间不能超过k

分析：DP+单调队列优化。

#include <queue> #include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAXN 0x3f3f3f3fusing namespace std;int n,m,k;int vl[105][10005],vr[105][10005],tl[105][10005],tr[105][10005],dp[105][10005],val[105][10005],dis[105][10005];int main(){while(~scanf("%d%d%d",&n,&m,&k) && n && m){memset(vl,0,sizeof(vl));memset(vr,0,sizeof(vr));memset(tl,0,sizeof(tl));memset(tr,0,sizeof(tr));n++;for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){  scanf("%d",&val[i][j]);  vl[i][j] = vl[i][j-1] + val[i][j];  }  for(int j = m;j;j--) vr[i][j-1] = vr[i][j] + val[i][j];}for(int i = 1;i <= n;i++){for(int j = 1;j <= m;j++){  scanf("%d",&dis[i][j]);  tl[i][j] = tl[i][j-1] + dis[i][j];  }  for(int j = m;j;j--) tr[i][j-1] = tr[i][j] + dis[i][j];}int ans = -MAXN;for(int i = 1;i <= n;i++){deque <int>lq;for(int j = 0;j <= m;j++){dp[i][j] = -MAXN; while(!lq.empty() && dp[i-1][j] - vl[i][j] >= dp[i-1][lq.back()] - vl[i][lq.back()]) lq.pop_back(); lq.push_back(j); while(!lq.empty() && tl[i][j] - tl[i][lq.front()] > k) lq.pop_front(); if(!lq.empty()) dp[i][j] = dp[i-1][lq.front()] - vl[i][lq.front()] + vl[i][j]; }deque <int>rq;for(int j = m;j >= 0;j--){while(!rq.empty() && dp[i-1][j] - vr[i][j] >= dp[i-1][rq.back()] - vr[i][rq.back()]) rq.pop_back();rq.push_back(j);while(!rq.empty() && tr[i][j] - tr[i][rq.front()] > k) rq.pop_front();    if(!rq.empty()) dp[i][j] = max(dp[i][j],dp[i-1][rq.front()] - vr[i][rq.front()] + vr[i][j]); if(i == n) ans = max(ans,dp[i][j]);}}cout<<ans<<endl;} }

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