BZOJ3445: [Usaco2014 Feb] Roadblock

数据范围很大，然而 n 很小

可以料想最短路树也是很小的

求最短路树，枚举最短路树上的树边，dijkstra 即可

---

代码：

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cstdio>
#include<queue>
using namespace std;

typedef long long ll;
const int MAXN = 255, MAXM = 250005;

struct EDGE{
	int nxt, to;
	ll val;
	EDGE(int NXT = 0, int TO = 0, ll VAL = 0ll) {nxt = NXT; to = TO; val = VAL;}
}edge[MAXM << 1];
int n, m, totedge = 1;
int head[MAXN], pre[MAXN], frm[MAXN];
bool vis[MAXN];
ll org, ans, dst[MAXN];
priority_queue<pair<ll,int> > q;

inline void add(int x, int y, ll v) {
	edge[++totedge] = EDGE(head[x], y, v);
	head[x] = totedge;
	return;
}
inline void dij() {
	for(int i = 1; i <= n; ++i) dst[i] = 2000000000000000;
	dst[1] = 0ll;
	q.push(make_pair(0, 1));
	while(!q.empty()) {
		int x = q.top().second; q.pop();
		if(vis[x]) continue;
		vis[x] = true;
		for(int i = head[x]; i; i = edge[i].nxt) {
			register int y = edge[i].to;
			if(dst[y] > dst[x] + edge[i].val) {
				dst[y] = dst[x] + edge[i].val;
				frm[y] = i;
				pre[y] = x;
				q.push(make_pair(-dst[y], y));
			}
		}
	}
	org = dst[n];
	return;
}
inline void dij2() {
	for(int i = 1; i <= n; ++i) dst[i] = 2000000000000000, vis[i] = false;
	dst[1] = 0ll;
	q.push(make_pair(0, 1));
	while(!q.empty()) {
		int x = q.top().second; q.pop();
		if(vis[x]) continue;
		vis[x] = true;
		for(int i = head[x]; i; i = edge[i].nxt) {
			register int y = edge[i].to;
			if(dst[y] > dst[x] + edge[i].val) {
				dst[y] = dst[x] + edge[i].val;
				q.push(make_pair(-dst[y], y));
			}
		}
	}
	return;
}
inline void work() {
	for(int y = 1; y <= n; ++y) {
		int x = pre[y], i = frm[y];
		edge[i].val <<= 1; edge[i ^ 1].val <<= 1;
		dij2();
		ans = max(ans, dst[n]);
		edge[i].val >>= 1; edge[i ^ 1].val >>= 1;
	}
	return;
}

int main() {
	scanf("%d%d", &n, &m);
	register int xx, yy;
	register ll vv;
	for(int i = 1; i <= m; ++i) {
		scanf("%d%d%lld", &xx, &yy, &vv);
		add(xx, yy, vv); add(yy, xx, vv);
	}
	dij();
	work();
	printf("%lld\n", ans - org);
	return 0;
}