【题解】绍兴一中-7.26-T3

最新推荐文章于 2024-11-22 19:41:17 发布

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本文分享了模拟赛T3题目的解题思路，通过优化的BFS算法，记录到达各点的最大体力值，减少冗余操作，实现从70分提升至90分的解题策略。

传送门
模拟赛的T3
没追求的我直接暴力BFS70分
很明显的需要记录状态 $v i s [u] [h p]$ ，表示是否出现过到点u，体力为hp这种情况
70分没毛病

比较秒的小优化思想是记录之前到过每个点的体力最大值，这样BFS时如果体力大于目标点之前最大体力才入队，vis数组就不用搞了
减少了好多冗余操作，这样一优化可以过90分
最后一个点需要2.25s，暂且当它过了吧

Code：

// luogu-judger-enable-o2
#include <bits/stdc++.h>
#define maxn 100010
using namespace std;
struct Edge{
	int to, next, len;
}edge[maxn << 1];
struct node{
	int u, hp, tim;
};
int n, m, M, num, head[maxn], R[maxn], maxhp[maxn];

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}

void addedge(int x, int y, int z){ edge[++num] = (Edge){ y, head[x], z }; head[x] = num; }

int main(){
	n = read(), m = read(), M = read();
	for (int i = 1; i <= n; ++i) R[i] = read();
	for (int i = 1; i <= m; ++i){
		int x = read(), y = read(), z = read();
		addedge(x, y, z); addedge(y, x, z);
	}
	maxhp[1] = M;
	queue <node> q; q.push((node) {1, M, 0});
	while (!q.empty()){
		node tmp = q.front(); q.pop();
		int u = tmp.u, hp = tmp.hp, tim = tmp.tim;
		for (int i = head[u]; i; i = edge[i].next){
			int v = edge[i].to, Hp = hp - edge[i].len;
			if (Hp <= 0) continue;
			Hp = min(Hp + R[v], M);
			if (Hp > maxhp[v]){
				maxhp[v] = Hp; q.push((node) {v, Hp, tim + 1});
				if (v == n){ printf("%d\n", tim + 1); return 0; }
			}
		}
	}
	puts("-1");
	return 0;
}

下面是std算法

我不想写了，直接来std代码：

#include <cstdio>
#include <algorithm>

using namespace std;

int edge[500000],next[500000],dist[500000],first[200000];
int b[200000],d[200000],g[10000000],h[10000000],p[200000],r[200000];
int i,k,m,n,x,y,z,head,tail,sum_edge;

int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for (i=1;i<=n;i++)
		scanf("%d",&r[i]);
	for (i=1;i<=m;i++)
	{
		scanf("%d%d%d",&x,&y,&z);
		sum_edge++,edge[sum_edge]=y,next[sum_edge]=first[x],dist[sum_edge]=z,first[x]=sum_edge;
		sum_edge++,edge[sum_edge]=x,next[sum_edge]=first[y],dist[sum_edge]=z,first[y]=sum_edge;
	}
	d[1]=p[1]=k;
	tail++,g[tail]=1,h[tail]=0;
	for (head=1;head<=tail;head++)
	{
		if (g[head]==n)
		{
			printf("%d\n",h[head]);
			return 0;
		}
		if (h[head]!=h[head-1])
			for (i=head;i<=tail;i++)
				d[g[i]]=p[g[i]],b[g[i]]=0;
		for (i=first[g[head]];i!=0;i=next[i])
			if ((d[g[head]]>dist[i]) && (min(d[g[head]]-dist[i]+r[edge[i]],k)>p[edge[i]]))
			{
				p[edge[i]]=min(d[g[head]]-dist[i]+r[edge[i]],k);
				if (! b[edge[i]])
					tail++,g[tail]=edge[i],h[tail]=h[head]+1,b[edge[i]]=1;
			}
	}
	printf("-1\n");
	return 0;
}