「LibreOJ NOIP Round #1」旅游路线

这道题真的好棒啊.. 不愧是WA爷出的题啊

题目大意

有$n$个点$m$条有向边，在每个点可以花$p_i$的钱加油至$min(C,c_i)$，每经过一条边会耗费$1$的油量

给出$T$个询问，每次询问从$s$出发，$q$块钱，走$d$的路程最多剩下多少钱

我好菜啊..

只能想到设$f_{i,q,c}$为走到$i$节点剩下$q$块钱还有$c$油量的最大路程来dp..

但是没法摆脱油量的限制

题解的做法很棒

就是设$f_{i,q}$为走到$i$节点并且在这里加油还剩下$q$块钱的最大路程，然后枚举下一次加油的地方来dp

那问题现在就转化为求从$i$到$j$不经过$min(C,c_i)$条边的最大路程

然后发现这个东西可以倍增求就很棒了

code

#include 
#include 
#include 
#include 
#define LL long long
using namespace std;
const LL Maxn = 110;
const LL Maxp = 100010;
const LL lg = 17;
const LL inf = (LL)1<<60;
LL f[Maxn][Maxn*Maxn];
LL g[lg][Maxn][Maxn];
LL w[Maxn][Maxn];
LL n, m, C, T;
LL p[Maxn], c[Maxn];
LL bit[Maxp];
LL A[Maxn], B[Maxn];
LL _max(LL x, LL y) { return x > y ? x : y; }
LL _min(LL x, LL y) { return x < y ? x : y; }
int main() {
	LL i, j, k;
	scanf("%lld%lld%lld%lld", &n, &m, &C, &T);
	for(i = 1; i <= n; i++) for(j = 1; j <= n; j++) g[0][i][j] = i == j ? 0 : -inf;
	for(i = 1; i <= n; i++) scanf("%lld%lld", &p[i], &c[i]);
	for(i = 1; i <= m; i++){
		LL x, y, d;
		scanf("%lld%lld%lld", &x, &y, &d);
		g[0][x][y] = _max(g[0][x][y], d);
	}
	for(LL x = 1; x < lg; x++){
		for(i = 1; i <= n; i++) for(j = 1; j <= n; j++) g[x][i][j] = -inf;
		for(k = 1; k <= n; k++){
			for(i = 1; i <= n; i++){
				for(j = 1; j <= n; j++) g[x][i][j] = _max(g[x][i][j], g[x-1][i][k]+g[x-1][k][j]);
			}
		}
	}
	for(i = 2; i <= 100000; i++) bit[i] = bit[i>>1]+1;
	for(i = 1; i <= n; i++){
		for(j = 1; j <= n; j++) A[j] = i == j ? 0 : -inf;
		LL u = _min(C, c[i]);
		for(int x = 0; x < lg; x++){
			if(u&(1<>1;
			if(f[s][mid] >= d){ ret = mid; R = mid-1; }
			else L = mid+1;
		}
		printf("%lld\n", q-ret);
	}
	return 0;
}