赛氪春季赛

oi大佬题解

B
建图！？
真没想到这个，图论见识的还是少
全源最短路
有时间再看看

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 300 + 5;
int a[N][N];
int d[N][N];

int main() {
    int n;
    cin >> n;
    memset(d, 63, sizeof d);
    for(int i = 0; i < n; ++ i) 
        for(int j = 0; j < n; ++ j)
            cin >> a[i][j];
    for(int i = 0; i < n; ++ i)
        for(int j = 0, k = 0;j < n; ++ j)
            if(a[i][j])
                d[i][j] = k++;

    for(int k = 0; k < n; ++ k)
        for(int i = 0; i < n; ++ i)
            for(int j = 0; j < n; ++ j)
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
    int ans = d[0][n-1] < 1 << 25 ? d[0][n-1] : -1;
    cout << ans << endl;
}

G
这码风真是一言难尽，有时间再补了，先把答案贴上

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
int fac[N], ifac[N];
int T[20];
typedef long long LL;

int pow(int a, int b) {
    LL res = 1;
    for(;b;b >>= 1, a = (LL) a * a % mod) 
        if (b & 1) 
            res = res * a % mod;
    return res;
}


int main() 
{
    int Max = 1000000;
    fac[0] = 1;
    for(int i = 1; i <= Max; ++ i) fac[i] = (LL)fac[i - 1] * i % mod;
    ifac[Max] = pow(fac[Max], mod - 2);
    for(int i = Max-1; i >= 0; -- i) ifac[i] = (LL)ifac[i + 1] * (i + 1) % mod;
    LL Sum = 0;
    int Count = 0;
    for(int i = 1; i <= 9; ++ i) {
        int temp;
        cin >> temp;
        T[i] = temp;
        Sum += i * temp;
        Count += T[i];
    } 
    Sum %= mod;
    LL part_I = 0;
    for(int i = 1; i <= Count; ++ i) part_I = (part_I * 10 + 1) % mod;
    LL part_mul = fac[Count - 1];
    LL part_div = 1;
    for(int i = 1; i<= 9; ++ i)  part_div = part_div * ifac[T[i]] % mod;
    // cout << Sum << ' ' << part_I << ' '  << part_mul << ' ' << part_div << endl;
    cout << (LL)Sum * part_I %mod * part_mul %mod * part_div % mod << endl;
    return 0; 
}

J题
是道比较难的dp，有时间再研究

	#include<cstdio>
	#include<cstring>
	#include<cmath>
	#include<algorithm>
	#include<iostream>
	#define ll long long
	using namespace std;
	ll f[2001][2001][3];
	ll ch[2001],vi[2001],ans;
	ll pre[2001],n;
	ll read(){
		ll x=0,w=1;char ch=getchar();
		while(ch>'9'||ch<'0'){if(ch=='-')w=-1;ch=getchar();}
		while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
		return x*w;
	}
	
	ll cal(ll l,ll r){
		return pre[l]+pre[n]-pre[r-1];
	}
	
	int main(){
		//freopen("10.in","r",stdin);
		//freopen("10.out","w",stdout);
		n=read();
		for(ll i=1;i<=n;i++)ch[i]=read(),ans+=ch[i];
		for(ll i=1;i<=n;i++)vi[i]=read();
		memset(f,63,sizeof(f));
	//	cout<<ans<<endl;
		for(ll i=1;i<=n;i++)f[i][i][0]=f[i][i][1]=0;
		for(ll i=1;i<=n;i++)pre[i]=pre[i-1]+vi[i];
		for(ll len=1;len<=n;len++){
			for(ll i=1;i<=n-len+1;i++){
				ll j=i+len-1;
				f[i][j][0]=min(f[i][j][0],f[i+1][j][0]+cal(i,j+1));//已经合并i+1,j,当前在i+1，准备向左到i 
				f[i][j][0]=min(f[i][j][0],f[i+1][j][1]+(j-i)*cal(i,j+1));//已经合并i+1,j,当前在j，准备向左到i 
				f[i][j][1]=min(f[i][j][1],f[i][j-1][0]+(j-i)*cal(i-1,j));//已经合并i,j-1,当前在i，准备向右到j 
				f[i][j][1]=min(f[i][j][1],f[i][j-1][1]+cal(i-1,j));//已经合并i,j-1,当前在j-1，准备向右到j
				//cout<<"f["<<i<<"]["<<j<<"]"<<" left="<<f[i][j][0]<<' '<<" right="<<f[i][j][1]<<endl;
			}
		}
	/*	for(ll i=1;i<=n;i++)
			for(ll j=1;j<=n;j++){
		//	cout<<"f["<<i<<"]["<<j<<"]"<<" left="<<f[i][j][0]<<' '<<" right="<<f[i][j][1]<<endl;
		}*/
	//	cout<<ans<<' '<<min(f[1][n][1],f[1][n][0])<<endl;
		printf("%lld\n",ans-min(f[1][n][1],f[1][n][0]));
		return 0;
	}

F题
小思维，题目数据有点怪，int居然不wa，而且这个开 ans 开 ll 就wa就离谱

#include<iostream>
using namespace std;
const int N = 200010;
typedef long long ll;
int n, m;
int ans[200010], sum, op, l, r, z;
int main()
{
	ios::sync_with_stdio(false);
	cin >> n >> m;
	while(m --)
	{
		cin >> op;
		if(op == 1)
		{
			cin >> l;
			cout << sum - ans[l] << endl;
		}
		else
		{
			cin >> l >> r >> z;
			for(int i = l;i <= r; ++i)
				ans[i] += (r - l + 1) * z;
			sum += (r - l + 1) * z;
		}
	}
	return 0;
}

C题
最小生成树
遇到必须添加的边的端点时， 就把这条必须加的边加进去

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N = 1e6 + 9;
ll n, m;
ll t;
ll f[N];
long long ans = 0;
struct node
{
	ll x, y, z;
}e[N];
bool cmp(node a, node b)
{
	return a.z < b.z;
}
/*ll find(ll x)
{
	return f[x] == x ? x : f[x] = find(f[x]);
}*/
ll find(ll x)
{
	ll r = x, j, k = x;
	while(r != f[r]) r = f[r]; // 寻找根节点 
	while(k != r)  
	{
		j = f[k];  // 暂存此时k的父节点 
		f[k] = r;  //  将k的父节点改为根节点 
		k = j;  //  k移到父节点   直至全改为根节点  
	}
	return r;
}
int main()
{
	cin >> n >> m;
	for(int i = 1; i <= m; ++i)
	{
		scanf("%lld%lld%lld", &e[i].x,&e[i].y,&e[i].z);	
	}	
	if(n == 1){
		cout << 0;return 0;
	}
	
	cin >> t;
	for(ll i = 0; i <= n; ++i) f[i] = i;
		
	ll x1 = e[t].x, x2 = e[t].y, x3 = e[t].z;
	
	sort(e+1,e+1+m,cmp);
	for(ll i = 1; i <= m; ++i)
	{
		if(e[i].x == x1 && e[i].y == x2 && e[i].z == x3) {
			t = i;break;
		}
	}
	//cout  << endl << t << endl;
	//f[e[t].x] = e[1].x;
	//f[e[t].y] = e[1].x;
	//ans += e[t].z;
	//cout << ans << endl;
	//if()
	bool q = 1;
	for(int i = 1; i <= m; ++i)
	{
		
		ll x = find(e[i].x);
		ll y = find(e[i].y);
		if(x == y) continue;
		//cout << e[i].x << " " << e[i].y  << " " << e[i].z << endl;
		if((e[i].x == e[t].x || e[i].x == e[t].y || e[i].y == e[t].x || e[i].y == e[t].y) && q)
		{
			//cout << "        \n";
			ans += e[t].z;
			f[find(e[t].x)] = x;
			f[find(e[t].y)] = x;
			q=0;
			if(e[i].x == e[t].x && e[i].y == e[t].y){
				continue;
			}
		}
		f[y] = x;
		ans += e[i].z;
		
		//cout << ans << endl;
	}
	cout << ans << endl;
	return 0;
}
/*
4 4
1 2 3
2 3 4
2 4 3
3 4 3
2

*/

zeal
直接上莫队算法即可。
另外题解