NOIP 2014飞扬的小鸟(DP优化)

题目链接  飞扬的小鸟

考场的70分暴力（实际只有50分因为数组开小了……）

考场代码（数组大小已修改）

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <functional>

 using namespace std;

 #define rep(i, a, b)    for (int i(a); i <= (b); ++i)
 #define dec(i, a, b)    for (int i(a); i >= (b); --i)
 #define LL        long long
 #define fi        first
 #define se        second

 const int INF =     <<    ;
 const int N =     +    ;
 const int M =     +    ;

 int dp[M][N], x[N], y[N], u[N], d[N];
 int n, m, q, ans, xn, ln, rn, _max, pos;
 bool c[N];

 int main(){

     scanf("%d %d %d ", &n, &m, &q);
     rep(i, , n - ) scanf("%d %d ", x + i, y + i);
     rep(i, , n) u[i] = m, d[i] = ;
     memset(c, false, sizeof c);
     rep(i, , q){
         scanf("%d %d %d ", &xn, &ln, &rn);
         c[xn] = true;
         u[xn] = rn - ;
         d[xn] = ln + ;
     }
     //rep(i, 0, n) printf("%d %d\n", d[i], u[i]);
     memset(dp, 0xff70, sizeof dp);_max = dp[][];
     rep(i, , m) dp[][i] = ;
     rep(i, , n - ){
         rep(j, d[i], u[i]){
             if (dp[i][j] >= _max) continue;
             int k = j - y[i];
             if (k >=  && k <= m) dp[i + ][k] = min(dp[i + ][k], dp[i][j]);
             k = j;int num = ;
             while (true){
                 k += x[i]; ++num;
                 if (k > m) k = m;
                 if (k >=  && k <= m) dp[i + ][k] = min(dp[i + ][k], dp[i][j] + num);
                 if (k == m) break;
             }
         }
     }

     ans = _max;
     rep(i, , m) ans = min(ans, dp[n][i]);
     if (ans < _max){
         printf("%d\n%d\n", , ans);return ;
     }

     bool flag = false;
     ans = ;
     dec(i, n, ) {rep(j, , m)  if (dp[i][j] < _max) { pos = i - ;flag = true; break;} if (flag) break;}
     dec(i, pos, ) if (c[i]) ++ans;
     printf("%d\n%d\n", , ans);

     return ;
 }

然后回过来想正解。

其实我们把很多时间都浪费在这里了：

     while (true){
         k += x[i]; ++num;
         if (k > m) k = m;
         if (k >=  && k <= m) dp[i + ][k] = min(dp[i + ][k], dp[i][j] + num);
         if (k == m) break;
     }

其实这一步可以转过来直接利用完全背包的性质优化一下。转移只要O（1）就可以了。

这道题细节还是很多的，比较容易写挂。

正解：

 #include <bits/stdc++.h>

 using namespace std;

 #define rep(i, a, b)    for (int i(a); i <= (b); ++i)
 #define dec(i, a, b)    for (int i(a); i >= (b); --i)

 const int INF =     <<    ;
 const int N =     +    ;
 const int M =     +    ;

 int dp[N][M], x[N], y[N], u[N], d[N];
 int n, m, q, ans, xn, ln, rn, _max, pos;
 bool c[N];

 int main(){

     scanf("%d %d %d ", &n, &m, &q);
     u[n] = m + ; d[n] = ;
     rep(i, , n - ) scanf("%d %d ", x + i, y + i), u[i] = m + , d[i] = ;

     rep(i, , q){
         scanf("%d %d %d ", &xn, &ln, &rn);
         u[xn] = rn;
         d[xn] = ln;
     }

     rep(i, , n + ) rep(j, , m + ) dp[i][j] = INF;

     rep(i, , m) dp[][i] = ;
     rep(i, , n){
         rep(j, x[i - ], m){
             dp[i][j] = min(dp[i][j], dp[i - ][j - x[i - ]] + );
             dp[i][j] = min(dp[i][j], dp[i][j - x[i - ]] + );
         }

         rep(j, m - x[i - ], m){
             dp[i][m] = min(dp[i][m], dp[i - ][j] + );
             dp[i][m] = min(dp[i][m], dp[i][j] + );
         }

         rep(j, d[i] + , u[i] - ){
             if (j + y[i - ] <= m) dp[i][j] = min(dp[i][j], dp[i - ][j + y[i - ]]);
         }

         rep(j, , d[i]) dp[i][j] = INF;
         rep(j, u[i], m) dp[i][j] = INF;
     }

     ans = INF;
     rep(i, , m) ans = min(ans, dp[n][i]);
     if (ans != INF) return , printf("%d\n%d\n", , ans);
     int t = q;
     dec(i, n, ){
         rep(j, , m) ans = min(ans, dp[i][j]);
         if (ans != INF) break;
         if (u[i] != m + ) --t;
     }

     printf("%d\n%d\n", , t);
     return ;
 }