[ZJOI2006]物流运输

Description

BZOJ1003

Solution

这个题一开始没有什么思路，看了黄学长的blog，才发现这个题是个很暴力的DP，设\(f[i]\)为前\(i\)天的花费，\(dist[i][j]\)为\([i,j]\)这几天的最短路径长度，则易得\(f[i] = min\{dist[1][i], f[j-1]+k+dist[j][i]\}\)。然后，dist数组竟然是暴力跑\(n^2\)遍最短路，复杂度\(O(n^2\cdot (m+n)logn)\)，虽然看起来过不了，但是由于会有很多点被ban掉，而且也跑不满，所以还是可以过掉的（再说时限10s怎么着也不会TLE吧）。

Code

#include <cstdio>
#include <queue>
#include <algorithm>
#include <cstring>

const int N = 25;
const int M = 110;
const int E = N*N;
const int INF = 1e9;
typedef long long LL;

LL dist[M][M], f[M];
int hd[N], to[E], nxt[E], w[E], cnt;
int ban[M][N];
int vis[N], dis[N];
int n, m, e, k;
struct node {
	int pos, w;
	node (int a = 0, int b = 0) : pos(a), w(b) {};
	bool operator < (const node &a) const {
		return w > a.w;
	}
};
std::priority_queue<node> q;

void adde(int x, int y, int z) {
	cnt++;
	to[cnt] = y;
	nxt[cnt] = hd[x];
	w[cnt] = z;
	hd[x] = cnt;
} 

int dij(int s, int t) {
	memset(vis, 0, sizeof vis);
	for (int i = 2; i <= m; ++i) dis[i] = INF;
	for (int i = s; i <= t; ++i) {
		for (int j = 2; j < m; ++j) {
			if (ban[i][j]) vis[j] = 1;
		}
	}
	q.push(node(1, 0));
	dis[1] = 0;
	node tmp;
	while (!q.empty()) {
		tmp = q.top(); q.pop();
		if (vis[tmp.pos]) continue;
		vis[tmp.pos] = 1;
		for (int i = hd[tmp.pos]; i; i = nxt[i]) if (!vis[to[i]]) {
			if (dis[to[i]] > tmp.w + w[i]) {
				dis[to[i]] = tmp.w + w[i];
				q.push(node(to[i], dis[to[i]]));
			}
		}
	}
	return dis[m];
}

int main() {
	scanf("%d%d%d%d", &n, &m, &k, &e);
	for (int i = 1, x, y, z; i <= e; ++i) {
		scanf("%d%d%d", &x, &y, &z);
		adde(x, y, z);
		adde(y, x, z);
	}
	int d;
	scanf("%d", &d);
	for (int i = 1, x, y, z; i <= d; ++i) {
		scanf("%d%d%d", &z, &x, &y);
		for (int j = x; j <= y; ++j) {
			ban[j][z] = 1; // mark
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = i; j <= n; ++j) {
			dist[i][j] = dij(i, j);
		}
	}
	f[0] = 0;
	for (int i = 1; i <= n; ++i) {
		f[i] = (LL)dist[1][i] * i;
		//printf("%d\n", f[i]);
		for (int j = 1; j <= i; ++j) {
			f[i] = std::min(f[i], f[j-1]+k+(LL)dist[j][i]*(i-j+1));
		}
	}
	printf("%lld\n", f[n]);
	return 0;
}

Note

mark的那个地方我竟然写成了ban[j][i]，还是要更仔细。其实这个数据范围是要开long long的，不过好像没有刻意卡，但还是要注意。