ICPC网络赛南京站 -分层图-数论

A题:题意求S=1 * 1!+2*2!+..+(n-1)*(n-1)!再%n的值

思路:

(n+1)!=n*n!+n!

1!+2!+3!+4!+...+(n+1)!=S+0!+1!+2!+3!+...n!

等式两边同时减去n!

S=n!-1

对于这个题

(n!-1)%n=n-1 // 由同余定理可以推出来

所以代码为

#include
using namespace std;

int main()
{
    int t;
    long long n;
    cin >> t;
    while(t--){
        scanf("%lld", &n);
        printf("%lld\n", n-1);
    }
    return 0;
}

 

有N个城市,M条单项路,起点是u,中点是v,权值是ci,你可以将不超过k条路的权值变成0.

现在计算从1到城市N最短路和的权值.

思路:分层图最短路

!!!我不是很懂分层图最短路写的思路,只能先把怎么实现说一说

有k条路免费 把图分解为k层 第i层有i条路免费

然后每一层都跑一遍最短路。

emmmm就是这样子

code:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#pragma warning(disable:4996)
using namespace std;
int s, t, n, m, k, d[100000 + 10][11];
const int INF = 0x3f3f3f3f;
typedef pair pii;
vector G[200000 + 10];
bool visit[100000 + 10][11];
struct Node
{
	int id;
	int level;
	int dis;
	Node(int id, int level, int dis) :id(id), level(level), dis(dis)
	{

	}
	bool operator <(const Node& X)const
	{
		return dis > X.dis;
	}
};
priority_queue pq;
int main(void)
{
	int T;
	cin >> T;
	while (T--)
	{
		int a, b, c;
		cin >> n >> m >> k;
		int s = 1, t = n;
		for (int i = 0; i <=n; i++)
			G[i].clear();
		for (int i = 0; i < m; i++)
		{
			scanf("%d%d%d", &a, &b, &c);
			G[a].push_back(make_pair(b, c));

		}
		memset(visit, 0, sizeof(visit));
		for (int i = 0; i <=n; i++)
			for (int j = 0; j <= k; j++)
				d[i][j] = INF;
		// visit[s][0]=true;
		d[s][0] = 0;//start from s,the zero level
		pq.push(Node(s, 0, 0));
		while (!pq.empty())
		{
			Node X = pq.top();
			pq.pop();
			int x = X.id;
			int lv = X.level;
			if (visit[x][lv] == true)
				continue;
			else
			{
				visit[x][lv] = true;
				vector::iterator it;
				for (it = G[x].begin(); it != G[x].end(); it++)
				{
					int y = (*it).first;
					int u = (*it).second;
					//the same level
					if (u + d[x][lv] < d[y][lv])
					{

						d[y][lv] = u + d[x][lv];
						pq.push(Node(y, lv, d[y][lv]));
					}
					//the +1 level
					if (lv < k)
						if (d[x][lv] < d[y][lv + 1])
						{
							d[y][lv + 1] = d[x][lv];
							pq.push(Node(y, lv + 1, d[y][lv + 1]));
						}
				}
			}
		}
		int ans = d[t][0];
		for (int i = 0; i <= k; i++)
		{
			ans = min(ans, d[t][i]);
		}
		cout << ans << endl;
	}
	return 0;
}

明早写一发dijkstra emmm先结束啦

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