BZOJ-1003 物流运输trans

对于每一个时间区间（共N^2个）我们都做一次最短路，然后DP。

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <fstream>
#include <iostream>
#include <queue>
#define rep(i, l, r) for(int i=l; i<=r; i++)
#define N 23
#define MAX 1<<30
#define clr(x, c) memset(x, c, sizeof(x))
using namespace std;
int read()
{
	int x=0, f=1; char ch=getchar();
	while (ch<'0' || ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	while (ch>='0' && ch<='9') { x=x*10+ch-'0'; ch=getchar(); }
	return x*f;
}
struct edge{int y, n, z;} e[N*N*2]; int fir[N], en;
struct data
{
	int x, d;
	data(int a, int b) : x(a), d(b) {}
	bool operator < (const data &k) const { return d>k.d; }
};
int n, m, v, p, d[N], k[123][123];
bool c[N][123];

inline void Add(int x, int y, int z)
{
	en++, e[en].y=y, e[en].z=z, e[en].n=fir[x], fir[x]=en;
	en++, e[en].y=x, e[en].z=z, e[en].n=fir[y], fir[y]=en;
}

int main()
{
	n=read(); m=read(); v=read(); clr(c, 1);
	p=read(); rep(i, 1, p) { int x=read(), y=read(), z=read(); Add(x, y, z); }
	p=read(); rep(o, 1, p) { int z=read(), x=read(), y=read(); rep(i, x, y) c[z][i]=false; }
	rep(s, 1, n) rep(t, s, n)
	{
		rep(i, 1, m) d[i]=-1;
		priority_queue<data> q; q.push(data(1, 0));
		while (d[m]<0 && !q.empty())
		{
			data a=q.top(); q.pop(); 
			int x=a.x, o=fir[x], y=e[o].y; 
			bool end=false; rep(i, s, t) if (!c[x][i]) end=true; if (end) continue;
			if (d[x]<0) d[x]=a.d; else continue;
			while (o) q.push(data(y, d[x]+e[o].z)), o=e[o].n, y=e[o].y;
		}
		if (d[m]<0) k[s][t]=MAX; else k[s][t]=d[m]*(t-s+1);
	}
	rep(i, 0, n-1) rep(j, 1, n-i) rep(o, j, j+i-1) k[j][j+i]=min(k[j][j+i], k[j][o]+k[o+1][j+i]+v);
	printf("%d\n", k[1][n]);
	return 0;
}