2013 whu暑假集训选拔#1
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一、已AC的题
D题（zoj3365）
开始看题后没发现什么思路，跑去想C题了，后来发现其他人瞬间把D过了，跑回去看题，才发现自己题目看错了，题目要求的是最终输出的是连续数字，及每个数与其对应的基数的差值，保留差值众数对应的数即可。
赛后来和天成讨论了一下，发现这个题目有个bug，输出的结果连续数只能上升么，如果是下降的话那么样例输出就是
2
5 4 3 2 1 0
而不是
3
3 4 5 6 7 8
题目讲的只是连续的数。。。
#include<iostream>#include<algorithm>using namespace std;#define M 50005int n;int a[M],b[M],c[M];int main(){int i,j;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++){scanf("%d",&a[i]);b[i]=a[i];a[i]-=i;c[i]=a[i];}sort(a+1,a+n+1);int temp=1;int max=0;int t=a[1];for(i=1;i<n;i++){if(a[i]==a[i+1])temp++;else{if(max<temp){max=max>temp?max:temp;t=a[i];}temp=1;}}if(a[n]==a[n-1]){max=max>temp?max:temp;t=a[i];}if(n==1)cout<<0<<endl;elsecout<<n-max<<endl;for(i=1;i<=n;i++){b[i]-=c[i]-t;}for(i=1;i<n;i++)cout<<b[i]<<" ";cout<<b[n]<<endl;}}二、赛后想想能过的题
H题（zoj3369）
一个比较简单的入门三维DP，之前写过一道类似的运武器的，可以对状态进行简化，使得空间复杂度为n^3
因为题目状态转移涉及四个变量，并且输出还要求输出路径
dp[i][j][k] 表示打第i个monster并且前面使用了j次fight，mana为k是的health
则有状态转移方程
dp[i+1][j+1][k] = max(a, b);
a = dp[i][j][k] - max(2*si - s0, 0) if(s0 + j >= si) //每次使用fight 有s++ 之前总共使用了j次
b = dp[i][j][k-mi] if(p + i - j >= pi && k >= mi)//每次使用enchant 有s++ 之前使用了 i-j次
最后就是逆向输出路径
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int MAXN = 52;int dp[MAXN][MAXN][MAXN];char path[MAXN][MAXN][MAXN];int ss[MAXN] , pp[MAXN] , mm[MAXN];int n, h, s0, p0, m0;char ans[MAXN];void gao(){ for(int i = 0; i < n; i++) for(int j = 0; j <= i; j++) for(int k = 0; k <= m0; k++) { if(dp[i][j][k] > 0) { if(s0 + j >= ss[i]) { int temp = max(2*ss[i] - (s0+j), 0); if(dp[i+1][j+1][k] < dp[i][j][k] - temp) { dp[i+1][j+1][k] = dp[i][j][k] - temp; path[i+1][j+1][k] = 'D'; } } if(p0 + i - j >= pp[i] && k >= mm[i]) { if(dp[i+1][j][k-mm[i]] < dp[i][j][k]) { dp[i+1][j][k - mm[i]] = dp[i][j][k]; path[i+1][j][k - mm[i]] = 'E'; } } } } bool isok = false; int pnum = -1 , mnum = -1; for(int j = 0; j <= n; j++) for(int k = 0; k <= m0; k++) { if(dp[n][j][k] > 0) { pnum = j; mnum = k; isok = true; break; } } if(isok) { for(int i = n-1; i >= 0; i--) { ans[i] = path[i+1][pnum][mnum]; if(ans[i] == 'E') mnum += mm[i]; else pnum--; } cout<<ans<<endl; } else cout<<"UNLUCKY"<<endl; }int main(){ while(scanf("%d%d%d%d%d", &n, &h, &s0, &p0, &m0) != EOF) { for(int i = 0; i < n; i++) scanf("%d%d%d", &ss[i], &pp[i], &mm[i]); for(int i = 0; i <= n; i++) for(int j = 0; j <= n; j++) for(int k = 0; k <= m0; k++) dp[i][j][k] = 0; dp[0][0][m0] = h; gao(); } return 0;}C题（zoj 3364）
开始总觉得是凸包，想了很久却没有想到处理办法，后来听文杰一点，焕然大悟。。
接下来就不吐槽了，思路是都是对的，想了两种方法计算黑白格子的数量，wa了这么久，就是一个初始化,a[1]和a[n+1]赋值的顺序问题，初始化函数写的是1到n，结果交了十几次才发现。。。
两个代码都贴上吧，一个1k多b，一个5k多b
思路，只计算水平方向的直线，往右走+，往左走-
1:
#include <iostream>#include <cmath>#include <cstring>#include <cstdio>#include <algorithm>#define LL long long#define abs(x) ((x)>0?(x):-(x))#define mset(a, b) memset(a, b, sizeof(a))#define rep(i, n) for(int i = 0; i < n; i++)using namespace std;const int MAXN = 50020;LL minx, miny;struct node{ LL x, y;}a[MAXN];int wwbb = -1;LL sum, bnum;int n;void init(){ minx = miny = 0xfffffffffffffffll; for(int i =1; i <= n; i++) { if(minx > a[i].x) minx = a[i].x; if(miny > a[i].y) miny = a[i].y; } LL t = abs(0 - minx) + abs(0 - miny); if(0 == t % 2) wwbb = 1;//黑 else wwbb = 0;//白 for(int i = 1; i <= n; i++) { a[i].x = a[i].x + 0 - minx; a[i].y = a[i].y + 0 - miny; }}LL calc(LL a, LL b){ LL ta = (a+1)/2, tb = (b+1)/2; LL ret = ta*tb; ta = a-ta; tb = b-tb; ret += ta*tb; if(wwbb == 0 && 1 == a % 2 && 1 == b % 2) { return ret - 1; }else return ret; }void gao(){ for (int i = 1; i <= n; i++) { if (a[i].y == a[i+1].y ){ sum += a[i+1].x*a[i+1].y; bnum += calc(a[i+1].x, a[i+1].y); sum -= a[i].x*a[i].y; bnum -= calc(a[i].x, a[i].y); } //cout<<bnum<<" "; }}int main(){ while(scanf("%d", &n) != EOF) { for(int i = 1; i <= n; i++) scanf("%lld%lld", &a[i].x, &a[i].y); init(); a[n+1] = a[1]; gao(); //cout<<endl; LL bans = 0, wans = 0; wans = abs(sum) - abs(bnum); bans = abs(bnum); printf("%lld %lld\n", bans, wans); sum = bnum = 0; } return 0;}2:
#include <iostream>#include <cmath>#include <cstring>#include <cstdio>#include <algorithm>#define LL long long#define abs(x) ((x)>0?(x):-(x))#define mset(a, b) memset(a, b, sizeof(a))#define rep(i, n) for(int i = 0; i < n; i++)using namespace std;const int MAXN = 50020;LL minx, miny;int wwbb = -1;struct node{ LL x, y; LL w, b;}a[MAXN];int n;void init(){ minx = miny = 0xfffffffffffffffll; for(int i =1; i <= n; i++) { if(minx > a[i].x) minx = a[i].x; if(miny > a[i].y) miny = a[i].y; } LL t = abs(0 - minx) + abs(0 - miny); if(0 == t % 2) wwbb = 1;//黑 else wwbb = 0;//白 for(int i = 1; i <= n; i++) { a[i].x = a[i].x + 0 - minx; a[i].y = a[i].y + 0 - miny; }}void gao(){ if(wwbb == 0)//白为底 { for(int i = 1; i <= n; i++) { if(a[i+1].x > a[i].x) { LL t = a[i+1].x - a[i].x; if(0 == t % 2) { a[i].w = -1 * ((t*a[i+1].y)/2); a[i].b = -1 * ((t*a[i+1].y)/2); } if(1 == t % 2) { if(0 == a[i].x % 2) { a[i].w = -1 * ((t-1)*a[i+1].y/2 + (a[i+1].y + 1)/2); a[i].b = -1 * ((t-1)*a[i+1].y/2 + (a[i+1].y)/2); } if(1 == a[i].x % 2) { a[i].b = -1 * ((t-1)*a[i+1].y/2 + (a[i+1].y + 1)/2); a[i].w = -1 * ((t-1)*a[i+1].y/2 + (a[i+1].y)/2); } } } if(a[i+1].x < a[i].x) { LL t = a[i].x - a[i+1].x; if(0 == t % 2) { a[i].w = ((t*a[i+1].y)/2); a[i].b = ((t*a[i+1].y)/2); } if(1 == t % 2) { if(0 == a[i+1].x % 2) { a[i].w = ((t-1)*a[i+1].y/2 + (a[i+1].y + 1)/2); a[i].b = ((t-1)*a[i+1].y/2 + (a[i+1].y)/2); } if(1 == a[i+1].x % 2) { a[i].b = ((t-1)*a[i+1].y/2 + (a[i+1].y + 1)/2); a[i].w = ((t-1)*a[i+1].y/2 + (a[i+1].y)/2); } } } } } if(wwbb == 1)//黑为底 { for(int i = 1; i <= n; i++) { if(a[i+1].x > a[i].x) { LL t = a[i+1].x - a[i].x; if(0 == t % 2) { a[i].b = -1 * ((t*a[i+1].y)/2); a[i].w = -1 * ((t*a[i+1].y)/2); } if(1 == t % 2) { if(0 == a[i].x % 2) { a[i].b = -1 * ((t-1)*a[i+1].y/2 + (a[i+1].y + 1)/2); a[i].w = -1 * ((t-1)*a[i+1].y/2 + (a[i+1].y)/2); } if(1 == a[i].x % 2) { a[i].w = -1 * ((t-1)*a[i+1].y/2 + (a[i+1].y + 1)/2); a[i].b = -1 * ((t-1)*a[i+1].y/2 + (a[i+1].y)/2); } } } if(a[i+1].x < a[i].x) { LL t = a[i].x - a[i+1].x; if(0 == t % 2) { a[i].b = ((t*a[i+1].y)/2); a[i].w = ((t*a[i+1].y)/2); } if(1 == t % 2) { if(0 == a[i+1].x % 2) { a[i].b = ((t-1)*a[i+1].y/2 + (a[i+1].y + 1)/2); a[i].w = ((t-1)*a[i+1].y/2 + (a[i+1].y)/2); } if(1 == a[i+1].x % 2) { a[i].w = ((t-1)*a[i+1].y/2 + (a[i+1].y + 1)/2); a[i].b = ((t-1)*a[i+1].y/2 + (a[i+1].y)/2); } } } } } }int main(){ while(scanf("%d", &n) != EOF) { for(int i = 1; i <= n; i++) scanf("%lld%lld", &a[i].x, &a[i].y); init(); a[n+1] = a[1]; gao(); LL bans = 0, wans = 0; for(int i = 1; i <= n; i++) { bans += a[i].b; a[i].b = 0; wans += a[i].w; a[i].w = 0; } printf("%lld %lld\n", bans, wans); //cout<<bans<<" "<<wans<<endl; } return 0;}三、能力范围外的题
A题
学了一下最小费用最大流，手敲了一遍模板，木有感觉。。看来图论得弄专题了。。
不过这题和一般的最小费用最大流不一样，把价格换算成赚的钱，及负数。。
#include<iostream>#include<cstdio>#include<queue>using namespace std;const int INF=999999999;int pre[105],head[105],d[105];struct hh{ int f,e,cap,cost,next;}edge[8050];int n,m,cent;void add1(int f,int e,int cap,int cost){ edge[cent].f=f;edge[cent].e=e; edge[cent].cap=cap;edge[cent].cost=cost; edge[cent].next=head[f]; head[f]=cent++;}void add(int f,int e,int cap,int cost){ add1(f,e,cap,cost); add1(e,f,0,-cost);}bool spfa(int s,int t){ queue<int>q; int u,v,cap,cost; bool in[105]={0}; for(int i=1;i<=n+1;i++) d[i]=INF; d[s]=0;in[s]=1; q.push(s); while(!q.empty()) { u=q.front();q.pop();in[u]=0; for(int i=head[u];i!=-1;i=edge[i].next) { v=edge[i].e;cap=edge[i].cap;cost=edge[i].cost; if(cap>0&&d[v]>d[u]+cost) { d[v]=d[u]+cost; pre[v]=i; if(!in[v]) { q.push(v);in[v]=1; } } } } return d[t]<0;}int EK(int s,int t){ int tp=INF; pre[s]=-1; for(int i=pre[t];i!=-1;i=pre[edge[i].f]) { tp=min(tp,edge[i].cap); } for(int i=pre[t];i!=-1;i=pre[edge[i].f]) { edge[i].cap-=tp; edge[i^1].cap+=tp; } return tp*d[t];}int main(){ int price,f,e,cap,cost,ans; while(scanf("%d%d",&n,&m)!=EOF) { cent=0;ans=0; for(int i=1;i<=n+1;i++)head[i]=-1; for(int i=2;i<=n;i++) { scanf("%d",&price); add(i,n+1,INF,-price); } for(int i=1;i<=m;i++) { scanf("%d%d%d%d",&f,&e,&cap,&cost); add(f,e,cap,cost); add(e,f,cap,cost); } while(spfa(1,n+1)) ans+=EK(1,n+1); printf("%d\n",-ans); } return 0;}E题
判断各个六边形联通块小于S时的大小，升序输出，用并查集维护每个联通块的大小
#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define N 50010#define V 500const int xx[6]={0,0,-1,-1,1,1} , yy[6]={1,-1,1,0,1,0} , _yy[6]={1,-1,0,-1,0,-1};int a[N],b[N],p[N],c[N] , num[2*V+10][2*V+10];bool land[2*V+10][2*V+10];int n,s,numn;int find(int x){ return x==p[x]?x:p[x]=find(p[x]);}int main(){ while(scanf("%d%d",&n,&s)!=EOF) { numn=0; memset(land,false,sizeof(land)); scanf("%d%d",&a[0],&b[0]); land[a[0]+V][b[0]+V]=true; num[a[0]+V][b[0]+V]=0; p[0]=0; c[0]=1; numn=1; for(int i=1; i<n; i++) { int x,y,ta,tb; scanf("%d%d",&ta,&tb); if(land[ta+V][tb+V]) continue; int sum,numr,r[10]; sum=0; numr=0; for(int k=0; k<6; k++) { x=ta+xx[k]; y=(ta&1)?tb+yy[k]:tb+_yy[k]; //if(x<-V || x>V || y<-V || y>V) continue; if(!land[x+V][y+V]) continue; int m=num[x+V][y+V]; int tmp=find(m); int j; for(j=0; j<numr; j++) if(r[j]==tmp) break; if(j==numr) r[numr++]=tmp; } for(int k=0; k<numr; k++) sum += c[r[k]]; if(sum<s) { a[numn] = ta; b[numn] = tb; land[ta+V][tb+V] = true; num[ta+V][tb+V] = numn; p[numn] = numn; c[numn] = sum+1; for(int k=0; k<numr; k++) p[r[k]]=numn; numn++; } } int Count=0,ans[N]; for(int k=0; k<numn; k++) if(p[k]==k) ans[Count++]=c[k]; sort(ans,ans+Count); printf("%d\n",Count); for(int k=0; k<Count; k++) { printf("%d",ans[k]); if(k==Count-1) printf("\n"); else printf(" "); } } return 0;}F题
整个分为两步，先求偏移量，N^2复杂度的枚举，接着就是一个动态规划求最大子矩阵
#include <stdio.h>#include <string.h>#include <stdlib.h>#define CY 135struct P{ int r, c, r2, c2, s, w, h;} ans;char ar[CY][CY], br[CY][CY];int C[CY], fro[CY], to[CY];int N, M, N2, M2;int main(void){ while (2 == scanf("%d%d", &N, &M)) { memset(ar, '@', sizeof(ar)); for (int i = 0; i < N; ++i) scanf("%s", ar[50 + i] + 50); scanf("%d%d", &N2, &M2); for (int i = 0; i < N2; ++i) scanf("%s", br[i]); ans.s = 0; for (int r = -N2 + 1; r < N; ++r) { for (int c = -M2 + 1; c < M; ++c) { memset(C, 0, sizeof(C)); for (int i = 0; i < N2; ++i) { for (int j = 0; j < M2; ++j) { if (ar[50 + i + r][50 + j + c] == br[i][j]) { C[j]++; } else { C[j] = 0; } } for (int j = 0; j < M2; ++j) { int k = j - 1; while (k >= 0 && C[k] >= C[j]) k = fro[k]; fro[j] = k; } for (int j = M2 - 1; j >= 0; --j) { int k = j + 1; while (k < M2 && C[k] >= C[j]) k = to[k]; to[j] = k; } for (int j = 0; j < M2; ++j) { int w = to[j] - fro[j] - 1; int h = C[j]; if (h * w > ans.s) { ans.s = h * w; ans.r2 = i - C[j] + 1; ans.c2 = fro[j] + 1; ans.r = r + ans.r2; ans.c = c + ans.c2; ans.h = h; ans.w = w; } } } } } if (ans.s == 0) { printf("0 0\n"); } else { printf("%d %d\n", ans.h, ans.w); printf("%d %d\n%d %d\n", ans.r + 1, ans.c + 1, ans.r2 + 1, ans.c2 + 1); } } return 0;}zoj1985
顺便把F题类似的zoj1985写写，很经典的迭代和枚举
#include <cstdio>using namespace std;long long a[100003];int l[100003],r[100003];int main(){ int n,i,j; long long MAX; while(scanf("%d",&n)&&n){ MAX=a[0]=a[n 1]=-1000000000; for(i=1;i<=n;i ){ scanf("%lld",a i); l[i]=1; j=i; while(a[j-l[j]]>=a[i]){ j-=l[j]; l[i] =l[j]; } } for(i=n;i>=1;i--){ r[i]=1; j=i; while(a[j r[j]]>=a[i]){ j =r[j]; r[i] =r[j]; } if((l[i] r[i]-1)*a[i]>MAX)MAX=(l[i] r[i]-1)*a[i]; } printf("%lld\n",MAX); } return 0;}I题
二分覆盖半径，判断是否为二分图，判断是否为二分图是用BFS，每个点加入队列中，如果flag[i]+flag[i+1] = 1，则满足是二分图。
#include <iostream>#include <cmath>#include <cstring>#include <cstdio>#include <queue>#include <algorithm>#define LL long long#define abs(x) ((x)>0?(x):-(x))#define mset(a, b) memset(a, b, sizeof(a))#define rep(i, n) for(int i = 0; i < n; i++)using namespace std;const int MAXN = 1220;const double eps = 1E-9;double d[MAXN][MAXN];int ans[MAXN];struct Point{ double x, y;}p[MAXN];double inline dis(const Point &l, const Point &r){ return sqrt((l.x-r.x)*(l.x-r.x) + (l.y-r.y)*(l.y-r.y));}int flag[MAXN];int n;bool isok(double rr){ mset(flag, -1); for(int i = 0; i < n; i++) { if(flag[i] > -1) continue; queue<int> q; q.push(i); flag[i] = 0; while(!q.empty()) { int now = q.front(); q.pop(); for(int j = 0; j < n; j++) { if(j == now || d[now][j] > rr) continue; if(flag[now] == flag[j]) return false; if(flag[j] == -1) { flag[j] = 1-flag[now]; q.push(j); } } } } for(int i = 0; i < n; i++) ans[i] = flag[i]+1; return true;}double bs(double l, double r){ while(l + eps < r) { double mid = (l + r)/2; if(isok(mid)) l = mid; else r = mid; } return l;}int main(){ while(~scanf("%d", &n) && n) { for(int i = 0; i < n; i++) { scanf("%lf%lf", &p[i].x, &p[i].y); for(int j = 0; j < i; j++) d[i][j] = d[j][i] = dis(p[i], p[j]); } double tmp = bs(0.0, 20000.0); printf("%.7lf\n", tmp*0.5); for(int i = 0; i < n; i++) printf("%d%c", ans[i], i == (n-1)?'\n':' '); } return 0;}- 2013 whu暑假集训选拔#1
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