20250814 OI 总结

Problem	T1	T2	T3	T4	T5	T6	all
want scores	100pts	100pts	60pts	40pts	60pts	0pts	360pts
scores	100pts	0pts	60pts	0pts	60pts	0pts	260pts

T3

爆零原因：输出制表符错了

正解：反向建图，这样每个点只会跑一次 dfs；

代码：（下次再也不用快读了）

#include <stdio.h>
#include <queue>
#include <vector>
#define ll long long
#define ull unsigned long long
#define int int
#define db double
using namespace std;
const int MAXN = 1e5 + 7;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n,m;
vector<int>e[MAXN];
int dis[MAXN];
int vis[MAXN];
int read(){
	int x = 0;
	char ch = getchar();
	while(ch < '0' || ch > '9'){ch = getchar();}
	while(ch >= '0' && ch <= '9'){x=x*10+(ch-'0');ch = getchar();}
	return x;
}
void dfs(int x,int root){
    if(vis[x]) return;
    vis[x] = root;
    dis[x] = 1;
    for(auto y : e[x]){
        if(!dis[y]){
            dfs(y,root);
        }
    }
}
int main(){
	n = read(),m = read();
	for(int i = 1;i <= m;i++){
		int x,y;
		x = read(),y = read();
		e[y].push_back(x);
	}
	for(int i = n;i >= 1;i--){
		dfs(i,i);
	}
    for(int i = 1;i <= n;i++)
        printf("%d ",vis[i]);
	return 0;
}

T4

死因：反转数组翻错了

思路：按照题目即可

方法总结：

1. 模拟题先拟好思路在写代码
2. 尽量一次写对，不然调试可是很爽的

代码：

#include <bits/stdc++.h>

using namespace std;
#define ll long long
#define ull unsigned long long
#define int int
#define db double
const int MAXN = 50 + 7;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int t,n,k;
char s[MAXN][MAXN],c[MAXN][MAXN];
char b[MAXN][MAXN],nm[MAXN][MAXN];
void fanzhuan(){
	for(int i = 1;i <= k;i++)
		for(int j = 1;j <= k;j++)
			b[j][k-i+1] = c[i][j];
	for(int i = 1;i <= k;i++)
		for(int j = 1;j <= k;j++)
			c[i][j] = b[i][j];
}
bool check(int x,int y){
	for(int i = 1,ii = x;i <= k;ii++,i++){
		for(int j = 1,jj = y;j <= k;j++,jj++){
			if(s[ii][jj] == '.' && c[i][j] == '*')
				return false;
		}
	}
	return true;
}
void draw(int x,int y){
	for(int i = 1,ii = x;i <= k;ii++,i++){
		for(int j = 1,jj = y;j <= k;j++,jj++){
			if(c[i][j] != '.'){
				nm[ii][jj] = '*';
			}
		}
	}
}
int main(){
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		for(int i = 1;i <= n;i++){
			for(int j = 1;j <= n;j++){
				nm[i][j] = '.';
			}
		}
		for(int i = 1;i <= n;i++) 
			scanf("%s",s[i]+1);
		scanf("%d",&k);
		for(int i = 1;i <= k;i++) 
			scanf("%s",c[i]+1);
		for(int i = 1;i <= n - k + 1;i++){
			for(int j = 1;j <= n - k + 1;j++){
				for(int l = 1;l <= 4;l++){
					fanzhuan();
					if(check(i,j)){
						draw(i,j);
					}
				}
			}
		}
		bool flag = 1;
		for(int i = 1;i <= n;i++){
			for(int j = 1;j <= n;j++){
				if(s[i][j] != nm[i][j]){
					flag = 0;
					break;
				}
			}	
		}
		printf("%s\n",(flag) ? "YES" : "NO");
	}	
	return 0;
}

T5

打表过样例，暴力出奇迹

死因：写完暴力没想出办法（原来写的二分，证明发现不行）

通过打表严谨分析，可以得出，只要求出顺序对的个数就好了。

代码：

#include <bits/stdc++.h>

using namespace std;
int n;
const int MAXN = 1e5 + 7;
int a[MAXN];
int c[MAXN]; 
long long ans = 0;
void qsort(int l,int r){
	if(l == r){
		return;
	}
	int mid = (l + r) / 2;
	int i = l,j = mid + 1,k = l;
	qsort(i,mid),qsort(mid+1,r);
	while(i <= mid && j <= r){
		if(a[i] <= a[j]){
			c[k++] = a[i++];
            ans += r - j + 1;
		}else{
			c[k++] = a[j++];
		}
	}
	while(i <= mid){
		c[k++] = a[i++];
	}
	while(j <= r){
		c[k++] = a[j++];
	}
	for(int c1 = l;c1 <= r;c1++){
		a[c1] = c[c1];
	}
}
int k;
int main()
{
	cin >> n >> k;
	for(int i = 1;i <= n;i++){
		cin >> a[i];
        a[i] -= k;
        a[i] += a[i-1];
	}
	qsort(0,n);
	cout << ans << endl;	
	return 0;
}

T6

死因：最小生成树不会（虽然今天学了）

怀疑这是原题搬过来的，点一下复制题目呃呃呃

代码：

#include <bits/stdc++.h>

using namespace std;
#define ll long long
const int INF  = 0x3f3f3f3f;
const int MAXH = 2e5 + 7;
const int mod = 1e9 + 7;
int n,m;
struct node{
	int x,y,w;
	bool operator < (const node &x) const{
		return w < x.w;
	}
}e[MAXH];
int fa[MAXH];
int c[MAXH];
int find(int x){
	if(fa[x] == x)
		return x;
	return fa[x] = find(fa[x]);
}
int kruskal(){
	for(int i = 1;i <= n;i++)
		fa[i] = i;
	sort(e+1,e+m+1);
	int tot = n,ans = 0;
	for(int i = 1;i <= m;i++){
		int x = find(e[i].x),y = find(e[i].y);
		if(x != y){
			ans += e[i].w;
			tot--;
			fa[x] = y;
		}
	}
    return ans;
}
signed main()
{
	cin >> n >> m;
    int s = INF;
    for(int i = 1;i <= n;i++){
        cin >> c[i];
        s = min(s,c[i]);
    }
	for(int i = 1;i <= m;i++){
		cin >> e[i].x >> e[i].y >> e[i].w;
        e[i].w += e[i].w;
        e[i].w += c[e[i].x] + c[e[i].y];
	}
	int ans = kruskal();
	cout << ans + s;
	return 0;
}