20250818 OI 总结

T2

前言：输出样例都给了贪心的方法了

正确思路：贪心，枚举，能还钱就还钱。

死因：贪心出错了（我的思路就是能还钱就还钱，但返回走下去）

代码：

#include <bits/stdc++.h>

using namespace std;

#define ll long long
#define ull unsigned long long
#define int int
const int MAXN = 1e5 + 7;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n;
int fuzhai_cnt = 0,shouyi_cnt = 0,pos,ans;
int a[MAXN];
signed main() {
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
        scanf("%d",&a[i]);
    for(int i = 1;i <= n;i++){
        if(a[i] < 0){
            if(!pos){
                pos = i;
            }
            fuzhai_cnt += a[i];
        }else{
            shouyi_cnt += a[i];
            if(shouyi_cnt >= (-(fuzhai_cnt)) && fuzhai_cnt != 0){
                shouyi_cnt += fuzhai_cnt;
                fuzhai_cnt = 0;
                ans += (i - pos) * 2;
                // cout << (i - pos) * 2 << endl;
                pos = 0;
            }
        }
        // cout << pos << endl;
    }
    printf("%d\n",ans+n);
    return 0;
}

T3

前言：类似题目（看起来好像是类似的） 嗅探器

死因：没清空

正确思路：dfs 即可

正解：

#include <bits/stdc++.h>

using namespace std;
#define ll long long
#define ull unsigned long long
#define int int
const int MAXN = 1e5 + 7;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
vector<int>e[MAXN];
int n,m;
int vis[MAXN],dis[MAXN];
void dfs(int x,int j){
	vis[x] = 1;
	for(auto y : e[x]){
		if(!vis[y]){
			if(y != j){
				dis[y] = dis[x] + 1;
				dfs(y,j);
			}
				
		}
	}
	vis[x] = 0;
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i = 1;i <= m;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	int s,t;
	int ans = 0;
	scanf("%d%d",&s,&t);
	
	for(int j = 1;j <= n;j++){
		memset(dis,-1,sizeof(dis));
		dfs(s,j);
		if(dis[t] == -1) ans++;
	}
	printf("%d\n",ans - 1);
	return 0;
}

歪解：

#include <bits/stdc++.h>

using namespace std;
#define ll long long
#define ull unsigned long long
#define int int
const int MAXN = 1e5 + 7;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
vector<int>e[MAXN];
int n,m;
int s,t,ans;
int dfn[MAXN],low[MAXN],times;
void tarjan(int u,int fa){
    low[u] = dfn[u] = ++times;
    bool flag = false;
    int son_ans = 0;
    for(auto v : e[u]){
        if(dfn[v]){
            low[u] = min(low[u],dfn[v]);
        }else{
            son_ans++;
            tarjan(v,u);
            low[u] = min(low[v],low[u]);
            if(!flag && fa != u && low[v] >= dfn[u] && dfn[t] >= dfn[v]){
                flag = 1;
                ans++;
            }
        }
    }
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i = 1;i <= m;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	scanf("%d%d",&s,&t);
	tarjan(s,s);
    printf("%d\n",ans);
	return 0;
}

总结：下次一定记得清空（已经抄了60次了）

T4

思路：找到循环节，每一个进行模拟即可（细节有一点多，还在调）

T5

思路：看 \(n\) 只有 6， \(dfs\) 启动（看上下左右的最小值求面积取一个 max）。

还在调试代码：

#include <bits/stdc++.h>

using namespace std;
#define ll long long
#define ull unsigned long long
#define int int
const int MAXN = 6 + 7;
const double INF = 1e9;
const int mod = 1e9 + 7;
const double PI = 3.1415926;
int n;
double sx,sy,ex,ey;
double x[MAXN],y[MAXN];
double dist[MAXN][MAXN],r[MAXN];
bool vis[MAXN];
double orz_dist(int i,int j){
	double res1 = (x[i] - x[j]) * (x[i] - x[j]);
	double res2 = (y[i] - y[j]) * (y[i] - y[j]);
	return sqrt(res1+res2);
}
double step = 0;
void dfs(int cur,double sum){
	if(cur > n){
		step = max(step,sum);
		return;
	}
	for(int i = 1;i <= n;i++){
		if(vis[i]) continue;
		double Rmin = INF;
		Rmin = min({abs(x[i]-ex),abs(sx-x[i]),abs(sy-y[i]),abs(ey-y[i])});
		for(int j = 1;j <= n;j++){
			if(!vis[j] || i == j) continue;
			Rmin = min(Rmin,max(0.0,dist[i][j]-r[j]));
		}
		vis[i] = 1;
		r[i] = Rmin;
		dfs(cur+1,sum + PI * r[i] * r[i]);
		vis[i] = 0;
		r[i] = 0;
	}
}
int main(){
	scanf("%d",&n);
	scanf("%lf%lf%lf%lf",&sx,&sy,&ex,&ey);
	if(sx > ex) swap(sx,sy),swap(ex,ey);
	if(sy > ey) swap(ex,ey);
	for(int i = 1;i <= n;i++)
		scanf("%lf%lf",&x[i],&y[i]);
	for(int i = 1;i <= n;i++)
		for(int j = 1;j <= n;j++)
			dist[i][j] = orz_dist(i,j);
	dfs(1,0.0);
	step = abs(sx-ex)*abs(sy-ey)-step;
	printf("%.0f\n",step);
	return 0;
}