Jzoj P5907 轻功___动态规划

题目大意：

给出一个长为$N$的序列，起点在$0$，有$K$种方式，每种可以花费$w_i$然后从某个位置$i$走到位置$i+C_i$，如果走到$i$用的是$j$方式，那么当你下一次走的时候方式不是$j$时，切换方式需要花费$W$，有$Q$个限制$a_i,b_i$，表示不能使用方式$a_i$经过位置$b_i$，问走到位置$N$的最少花费。

$N<=500，K<=100$
$Q<=50000，W<=1e7$

分析：

设$f[i][j]$表示走到位置$i$时方式为$j$的最少花费，
$f[i][j]=min(f[i-C_k][k]+w_i+((k=j)?0:W))$
时间复杂度:$O(N{K}^2)$

代码：

#include 
#include 
#include 
#include 
#include 
#include 
#define inf 0x7fffffff
#define N 505
#define M 105

using namespace std;

typedef long long ll;

struct Node { int len; ll cost; }a[M];
int sum[M][N], n, m, Q;
ll f[N][N], W;

int read(int &x)
{
	int f = 1; x = 0; char s = getchar();
	while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
	while (s >= '0' && s <= '9') { x = x * 10 + (s - '0'); s = getchar(); }
	x = x * f;
}

void write(ll x)
{
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}

int main()
{
    freopen("qinggong.in", "r", stdin);
    freopen("qinggong.out", "w", stdout);
	read(n); read(m); ++n; scanf("%lld", &W);
    for (int i = 1; i <= m; i++) read(a[i].len), scanf("%lld", &a[i].cost);
    read(Q);
    while (Q--)
    {
           int x, y; read(x); read(y);
           sum[y][x + 1] = 1;
	}
	for (int i = 1; i <= m; i++) 
	     for (int j = 1; j <= n; j++) sum[i][j] = sum[i][j] + sum[i][j - 1];	
	for (int i = 2; i <= n; i++) 
	     for (int j = 1; j <= m; j++) f[i][j] = inf;
	     
	for (int i = 2; i <= n; i++)
		 for (int j = 1; j <= m; j++)
		      if (i > a[j].len)
	              if (sum[j][i] - sum[j][i - a[j].len - 1] == 0)
	              {
	          	      for (int k = 1; k <= m; k++)
				            if (k != j) f[i][j] = min(f[i][j], f[i - a[j].len][k] + W + a[j].cost);
					               else f[i][j] = min(f[i][j], f[i - a[j].len][k] + a[j].cost);	
				  }
		  
	ll ans = inf;
	for (int i = 1; i <= m; i++) ans = min(ans, f[n][i]);
	if (ans == inf) printf("-1\n");
	         else { write(ans); printf("\n"); }
	return 0;	
}