dp转移方程很容易想 dp[i][j] = min{dp[i - 1][k] + abs(pos[i][j] -pos[i - 1][j]) + cost[i][j]}
n行m列 每次转移扫描m次 共n*m*m 为O(10^7) 1500ms，可以暴力一试。姿势不对就会TLE
其实加上个内联函数求绝对值，同时赋值时候不使用min(a, b) 用G++交 就可以水过
正解是：因为每个转移都是从上一层所有的状态开始转移，将上一层的状态根据pos排序
对当前状态的pos进行讨论，其坐标轴左边的点dp[i][j] = dp[i - 1][k] - pos][i - 1][k]+ pos[i][j] + cost[i][j]
对其坐标轴右边的点便是 dp[i][j] = dp[i - 1][k]+ pos][i - 1][k] - pos[i][j] + cost[i][j]， pos][i][j] 和 cost[i][j]是常数。
维护一个lans[i],表示上一层0 ~ i位置的dp[i - 1][k] - pos][i - 1][k]最小值； 和一个rans[i]，表示上一层i ~ (m - 1)位置的dp[i - 1][k] + pos][i - 1][k]最小值
二分当前状态的pos,若为p，比较左边lans[p - 1]与右边lans[p]的最小值即可
//#pragma comment(linker, "/STACK:102400000,102400000")
//HEAD
#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <cstdlib>
using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
//STL
#define PB push_back
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)
#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)
#define ALL(c) (c).begin(), (c).end()
#define SZ(V) (int)V.size()
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)
#define sqr(x) x * x
typedef long long LL;
typedef vector <int> VI;
typedef unsigned long long ULL;
const double eps = 1e-10;
const LL MOD = 1e9 + 7;
using namespace std;
const int maxn = 1010;
const int INF = 0x3f3f3f3f;
int dp[55][maxn], n, m, x;
int lm[maxn], rm[maxn], lans[maxn], rans[maxn];
struct Node{
int pos, cost;
int lval, rval;
bool operator <(const Node& a) const
{
return pos < a.pos;
}
}a[55][maxn];
void init(int u)
{
REP(j, m)
{
a[u][j].lval = dp[u][j] - a[u][j].pos;
a[u][j].rval = dp[u][j] + a[u][j].pos;
}
sort(a[u], a[u] + m);
CLR(lm, INF), CLR(rm, INF);
int s = 0, e = 0;
lans[0] = lm[0] = a[u][0].lval, rans[m - 1] = rm[0] = a[u][m - 1].rval;
FF(j, 1, m)
{
Node &v = a[u][j];
if (v.lval > lm[e]) lm[++e] = v.lval;
else
{
while (e >= 0 && v.lval < lm[e]) --e;
lm[++e] = v.lval;
}
lans[j] = lm[0];
}
s = e = 0;
FED(j, m - 2, 0)
{
Node &v = a[u][j];
if (v.rval > rm[e]) rm[++e] = v.rval;
else
{
while (e >= 0 && v.rval < rm[e]) --e;
rm[++e] = v.rval;
}
rans[j] = rm[0];
}
}
int main()
{
int T;
RI(T);
while (T--)
{
RIII(n, m, x);
REP(i, n) REP(j, m) RI(a[i][j].pos);
REP(i, n) REP(j, m) RI(a[i][j].cost);
REP(j, m) dp[0][j] = abs(x - a[0][j].pos) + a[0][j].cost;
FE(i, 1, n - 1)
{
init(i - 1);
REP(j, m)
{
int p = lower_bound(a[i - 1], a[i - 1] + m, a[i][j]) - a[i - 1];
// cout << "position "<< p << endl;
int lmin = INF, rmin = INF;
if (p) lmin = lans[p - 1] + a[i][j].pos + a[i][j].cost;
if (p < m) rmin = rans[p] - a[i][j].pos + a[i][j].cost;
// cout << "rans[p]: " << rans[p] << endl;
// cout << "lmin: " <<lmin << "rmin: " << rmin << endl;
dp[i][j] = min(lmin, rmin);
}
}
int ans = INF;
REP(j, m) ans = min(dp[n - 1][j], ans);
cout << ans << endl;
}
return 0;
}
//#pragma comment(linker, "/STACK:102400000,102400000")
//HEAD
#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <cstdlib>
using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
//STL
#define PB push_back
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)
#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)
#define ALL(c) (c).begin(), (c).end()
#define SZ(V) (int)V.size()
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)
#define sqr(x) x * x
typedef long long LL;
typedef vector <int> VI;
typedef unsigned long long ULL;
const double eps = 1e-10;
const LL MOD = 1e9 + 7;
using namespace std;
const int maxn = 1010;
const int INF = 0x3f3f3f3f;
inline int ABS(int x) {if (x < 0) return -x; return x; }
int pos[55][maxn], cost[55][maxn];
int dp[55][maxn];
int main()
{
int T, n, m, x;
RI(T);
while (T--)
{
RIII(n, m, x);
REP(i, n) REP(j, m) RI(pos[i][j]);
REP(i, n) REP(j, m) RI(cost[i][j]);
REP(i, m)
dp[0][i] = ABS(x - pos[0][i]) + cost[0][i];
FE(i, 1, n - 1)
REP(j, m)
{
int t = INF;
REP(k, m)
{
// dp[i][j] = min(dp[i][j], dp[i - 1][k] + ABS(pos[i][j] - pos[i - 1][k]));
int x = dp[i - 1][k] + ABS(pos[i][j] - pos[i - 1][k]);
if (x < t)
t = x;
}
dp[i][j] = t + cost[i][j];
}
int ans = INF;
REP(i, m) ans = min(ans, dp[n - 1][i]);
WI(ans);
}
return 0;
}hdu4362 dp + 单调队列优化,布布扣,bubuko.com
原文地址：http://blog.csdn.net/colin_27/article/details/37724663
