洛谷 P1821 [USACO07FEB]银牛派对Silver Cow Party

银牛派对

正向建图+反向建图， 两边跑dijkstra，然后将结果相加即可。

反向建图以及双向建图的做法是学习图论的必备思想。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
//Mystery_Sky
//
#define maxn 1000010
#define maxm 5000050
#define INF 0x3f3f3f3f
struct Edge{
	int next;
	int w;
	int to;
}edge1[maxn];
Edge edge2[maxn];
int n, m, X;
int head1[maxn], head2[maxn], cnt1, cnt2;
int vis1[maxn], vis2[maxn], dis1[maxn], dis2[maxn];

inline void add_edge1(int u, int v, int w)
{
	edge1[++cnt1].to = v;
	edge1[cnt1].next = head1[u];
	edge1[cnt1].w = w;
	head1[u] = cnt1;
}

inline void add_edge2(int u, int v, int w)
{
	edge2[++cnt2].to = v;
	edge2[cnt2].next = head2[u];
	edge2[cnt2].w = w;
	head2[u] = cnt2;
}

struct node{
	int dis;
	int pos;
	inline bool operator <(const node &x) const
	{
		return x.dis < dis;
	}
};
priority_queue <node> q1;
priority_queue <node> q2;

inline void dijkstra1()
{
	dis1[X] = 0;
	q1.push((node) {0, X});
	while(!q1.empty()) {
		node top = q1.top();
		q1.pop();
		int x = top.pos;
		if(vis1[x]) continue;
		vis1[x] = 1;
		for(int i = head1[x]; i; i = edge1[i].next) {
			int y = edge1[i].to;
			if(dis1[y] > dis1[x] + edge1[i].w) {
				dis1[y] = dis1[x] + edge1[i].w;
				if(!vis1[y]) q1.push((node) {dis1[y], y});
			}
		}
	}
}

inline void dijkstra2()
{
	dis2[X] = 0;
	q2.push((node) {0, X});
	while(!q2.empty()) {
		node top = q2.top();
		q2.pop();
		int x = top.pos;
		if(vis2[x]) continue;
		vis2[x] = 1;
		for(int i = head2[x]; i; i = edge2[i].next) {
			int y = edge2[i].to;
			if(dis2[y] > dis2[x] + edge2[i].w) {
				dis2[y] = dis2[x] + edge2[i].w;
				if(!vis2[y]) q2.push((node) {dis2[y], y});
			}
		}
	}
}
int ans = 0;
int main() {
	scanf("%d%d%d", &n, &m, &X);
	int u, v, w;
	memset(dis1, INF, sizeof(dis1));
	memset(dis2, INF, sizeof(dis2));
	for(int i = 1; i <= m; i++) {
		scanf("%d%d%d", &u, &v, &w);
		add_edge1(u, v, w);
		add_edge2(v, u, w);
	}
	dijkstra1();
	dijkstra2();
	for(int i = 1; i <= n; i++) {
		if(i == X) continue;
		if(ans < dis1[i] + dis2[i]) ans = dis1[i] + dis2[i];
	}
	printf("%d\n", ans);
	return 0;
}