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：
原文：http://www.cnblogs.com/Aragaki/p/7197786.html
