soj 4511 dp差值

Problem:
Analyse:
- 差值dp,最大限度的减少信息容易,而有最低的维度.
- 这里要想象有两个任务堆,左边一个,右边一个,然后往上面添加任务.
- 这个题与前面的积木堆得不同是,任务的执行顺序的限制,这需要在dp转移的时候结合实际来考虑.
- 定义dp[i][j]:前i个物品形成左堆和右堆得差值为j的序列的最小时间 .
- 转的核心就是本机器里的任务要执行完了再叠加,内存给的比较小,用滚动数组实现.
Get:
差值dp的经典特点是两堆,然后维护该差值下的最值即可.
Code:

/**********************jibancanyang************************** *Author*        :jibancanyang *Created Time*  : 日  5/15 15:00:03 2016***********************[e-mail]**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;vector<int> vi;#define pr(x) cout << #x << ": " << x << "  " #define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a));#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++) const int mod = int(1e9) + 7, INF = 0x3f3f3f3f;const int maxn = 6e3 + 13;int dp[2][maxn * 2];int a[2009], b[2016];int main(void){#ifdef LOCAL    freopen("/Users/zhaoyang/in.txt", "r", stdin);    //freopen("/Users/zhaoyang/out.txt", "w", stdout);#endif    int T; sa(T);    while (T--) {        int n, sum = 0; sa(n);        for (int i = 0; i < n; i++) sa(a[i]), sa(b[i]), sum = max(sum, max(a[i], b[i]));         for (int i = -sum; i <= sum; i++) dp[0][i + maxn] = INF;        dp[0][0 + maxn] = 0;        for (int i = 0; i < n; i++) {            for (int j = -sum; j <= sum; j++) dp[(i + 1) % 2][j + maxn] = INF;            for (int j = -sum; j <= sum; j++) {                if (dp[i % 2][j + maxn] != INF) {                    if (j <= 0) {                         dp[(i + 1) % 2][j + a[i] + maxn] = min(dp[(i + 1) % 2][j + a[i] + maxn] ,dp[i % 2][j + maxn] + ((j + a[i]) < 0 ? 0 : (j + a[i])));                         dp[(i + 1) % 2][-b[i] + maxn] = min(dp[(i + 1) % 2][-b[i] + maxn], dp[i % 2][j + maxn] + b[i]);                    } else {                         dp[(i + 1) % 2][a[i] + maxn] = min(dp[(i + 1) % 2][a[i] + maxn], dp[i % 2][j + maxn] + a[i]);                         dp[(i + 1) % 2][j - b[i] + maxn] = min(dp[(i + 1) % 2][j - b[i] + maxn], dp[i % 2][j + maxn] + ((b[i] - j) < 0 ? 0 : (b[i] - j)));                    }                }            }        }        int ans = INF;        for (int i = -sum; i <= sum; i++)   ans = min(ans, dp[n % 2][i + maxn] );        pri(ans);    }    return 0;}