B：读懂题意模拟
#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#define foror(i,a,b) for(i=a;i<b;i++)
#define foror2(i,a,b) for(i=a;i>b;i--)
#define EPS 1.0e-6
#define PI acos(-1.0)
#define INF 3000000000
#define MOD 1000000009
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
//using ll = long long;
//using ull= unsigned long long;
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
typedef long long ll;
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
int n;
int m1,m2,y1,y2;
cin >> n;
double ans;
while(n--)
{
scanf("%d %d %d %d",&m1,&y1,&m2,&y2);
//cout<< a << b << c << d<<endl;
if(y1==y2)
ans=(m2-m1)/2.0/(13-m1)+EPS;
else
ans=0.5+(y2-y1-1)+1.0/12*(m2-1)+EPS;
printf("%.4lf\n",ans);
}
return 0;
}
C：给你一个起点和终点 每个点有补给但是只能拿一次 每条路有经过的cost 问你能不能到终点 能的话最少cost多少 DFS + 剪枝
(对于一个走过点k,如果有必要再走一次,那么一定是走过k后在k点的最大弹药数增加了.否则一定没有必要再走. 记录经过每个点的最大弹药数,对dfs进行剪枝.)
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <map>
using namespace std;
map<string, int> pos;
struct Edge {
int v, c, next;
} edge[3000];
int head[3000], cnt;
int vis[3000], amm[3000], aim[3000],sum[3000];
int cs, n, m, ans;
string s, t;
void addedge (int u, int v, int c) {
edge[++cnt].v = v, edge[cnt].c = c;
edge[cnt].next = head[u];
head[u] = cnt;
}
void dfs (int x, int s, int k) {
if(k>=ans) return;
if (aim[x]) {
ans = min (ans, k);
return ;
}
if (!vis[x]) s += amm[x], vis[x] = 1;
if(vis[x]&&sum[x]>=s) return;
sum[x]=max(sum[x],s);
for (int i = head[x]; i; i = edge[i].next) {
int v = edge[i].v, c = edge[i].c;
if (s >= c) dfs (v, s - c, k + c);
}
vis[x] = 0;
}
int main() {
scanf ("%d", &cs);
while (cs--) {
pos.clear();
memset (head, 0, sizeof head);
memset(sum,0,sizeof sum);
memset(vis,0,sizeof vis);
cnt = 1, ans = 0x7fffffff;
scanf ("%d %d", &n, &m);
for (int i = 1; i <= n; i++) {
cin >> s;
pos[s] = i;
cin >> amm[i] >> s;
if (s[0] == ‘y‘) aim[i] = 1;
else aim[i] = 0;
}
for (int i = 1, x; i <= m; i++) {
cin >> s >> t >> x;
int u = pos[s], v = pos[t];
addedge (u, v, x);
addedge (v, u, x);
}
dfs (1, 0, 0);
if (ans != 0x7fffffff) printf ("%d\n", ans);
else puts ("No safe path");
}
}
D
E:如果有 就先去掉特殊的两个 然后DFS
#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#define foror(i,a,b) for(i=a;i<b;i++)
#define foror2(i,a,b) for(i=a;i>b;i--)
#define EPS 1.0e-6
#define PI acos(-1.0)
#define INF 3000000000
#define MOD 1000000009
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
//using ll = long long;
//using ull= unsigned long long;
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
typedef long long ll;
int num[10];
int flag=0;
void dfs(int number)
{
if(number==4||flag==1)
{
flag=1;
return ;
}
for(int i=1;i<=7;i++)
{
if(num[i]>=1&&num[i+1]>=1&&num[i+2]>=1)
{
num[i]-=1;
num[i+1]-=1;
num[i+2]-=1;
dfs(number+1);
num[i]+=1;
num[i+1]+=1;
num[i+2]+=1;
}
}
for(int i=1;i<=9;i++)
{
if(num[i]>=3)
{
num[i]-=3;
dfs(number+1);
num[i]+=3;
}
}
}
void init()
{
for(int i=1;i<=9;i++)
{
if(num[i]<2)
continue;
num[i]-=2;
dfs(0);
num[i]+=2;
if(flag==1)
break;
}
}
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
int n;
cin >> n;
while(n--)
{
flag=0;
mem(num,0);
int a[20];
for(int i=1;i<=14;i++)
{
scanf("%d",&a[i]);
num[a[i]]++;
}
init();
if(flag==1)
printf("Vulnerable\n");
else
printf("Immune\n");
}
return 0;
}
F:BFS 注意！！一次outbreak只能在BFS中的每个点发生一次
#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#define foror(i,a,b) for(i=a;i<b;i++)
#define foror2(i,a,b) for(i=a;i>b;i--)
#define EPS 1.0e-6
#define PI acos(-1.0)
#define INF 3000000000
#define MOD 1000000009
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
//using ll = long long;
//using ull= unsigned long long;
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
typedef long long ll;
char s[110][110];
char visit[110][110];
queue<pair<int,int> > que;
int dir[4][2];
int n,m;
void bfs()
{
for(int i=0;i<=101;i++)
mem(visit[i],0);
pair<int,int> a;
while(!que.empty())
{
a=que.front();
que.pop();
int r=a.first;
int c=a.second;
if(s[r][c]==‘D‘)
{
for(int i=0;i<4;i++)
{
int r1=r+dir[i][0];
int c1=c+dir[i][1];
if(r1>=0&&r1<n&&c1>=0&&c1<m&&s[r1][c1]!=‘X‘)
{
if(s[r1][c1]==‘D‘&&!visit[r1][c1])
{
pair<int,int> b;
b.first=r1,b.second=c1;
que.push(b);
visit[r1][c1]=1;
}
else if(s[r1][c1]>=‘A‘&&s[r1][c1]<‘D‘)
{
s[r1][c1]=s[r1][c1]+1;
}
}
}
visit[r][c]=1;
}
else if(s[r][c]>=‘A‘&&s[r][c]<‘D‘)
{
s[r][c]=s[r][c]+1;
}
}
}
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
dir[0][0]=-1,dir[0][1]=0;
dir[1][0]=1,dir[1][1]=0;
dir[2][0]=0,dir[2][1]=1;
dir[3][0]=0,dir[3][1]=-1;
int t;
cin >> t;
while(t--)
{
for(int i=0;i<=101;i++)
mem(visit[i],0);
int r,c;
scanf("%d %d",&m,&n);
for(int i=0;i<n;i++)
scanf("%s",s[i]);
int time;
scanf("%d",&time);
while(time--)
{
scanf("%d %d",&c,&r);
pair<int,int> a;
a.first=r,a.second=c;
que.push(a);
bfs();
/*for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
printf("%c",s[i][j]);
printf("\n");
}
cout<<endl;*/
}
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
printf("%c",s[i][j]);
printf("\n");
}
}
return 0;
}
G：水
#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#define foror(i,a,b) for(i=a;i<b;i++)
#define foror2(i,a,b) for(i=a;i>b;i--)
#define EPS 1.0e-8
#define PI acos(-1.0)
#define INF 3000000000
#define MOD 1000000009
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
//using ll = long long;
//using ull= unsigned long long;
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
typedef long long ll;
int a[25];
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
int n;
cin >> n;
while(n--)
{
for(int i=0;i<20;i++)
cin>>a[i];
for(int i=19;i>0;i--)
{
if(i!=0)
a[i-1]+=a[i]/2;
a[i]=a[i]%2;
}
cout<<a[0];
for(int i=1;i<20;i++)
cout<<" "<<a[i];
cout<<endl;
}
return 0;
}
F：以分号为界 每个一行 按照题意输出 substr的应用
#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#define foror(i,a,b) for(i=a;i<b;i++)
#define foror2(i,a,b) for(i=a;i>b;i--)
#define EPS 1.0e-8
#define PI acos(-1.0)
#define INF 3000000000
#define MOD 1000000009
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
//using ll = long long;
//using ull= unsigned long long;
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
typedef long long ll;
string a,b;
int now=0;
int pop=0;
void doit()
{
int cur=0;
string ans;
for(int i=0;i<b.size();i++)
{
if(b[i]==‘;‘)
{
ans=b.substr(cur,i-cur+1);
if(pop==0)
printf("%d: ",now-1);
else
printf("%d: ",now);
cout<<ans<<endl;
cur=i+1;
}
}
b=b.substr(cur,b.size()-cur+1);
}
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
int t;
cin >> t;
now=0;
b=a="";
while(getline(cin,a))
{
now++;
if(a=="END OF CASE")
{
pop=1;
now=0;
b="";
continue;
}
b+=a;
doit();
}
return 0;
}
J：
时间： 2024-10-25 08:39:17