2014ACM/ICPC亚洲区北京站现场赛(HDU 5111,5115,5119,5120,5122)

HDU 5112

水题，注意取绝对值和浮点数

#include<stdio.h>#include<algorithm>#include<math.h>using namespace std;struct Point {    int t;    int x;};int cmp(Point a,Point b) {    return a.t<b.t;}Point p[10005];int main() {    #ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    int T,n;    int cas=1;    scanf("%d",&T);    while(T--) {        scanf("%d",&n);        int i;        for(i=0;i<n;i++) {            scanf("%d%d",&p[i].t,&p[i].x);        }        sort(p,p+n,cmp);        double max=-1,v;        for(i=1;i<n;i++) {            v=double(1.0*abs(p[i].x-p[i-1].x)/(p[i].t-p[i-1].t*1.0));            if (v>max) max=v;        }        printf("Case #%d: %.2lf\n",cas++,max);    }}

HDU 5122

这个题卡了好久，刚开始思路就错了。以为是之前想过的冒泡外层循环的次数，直接求最大逆序数。WA了4个小时。最后才发现题意就理解错了。

树状数组，每次移动最大数字，那它位置后的数字都前移一位。每次看以后每次在判断次数。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define lowbit(x) (x&-x)#define maxn 1100000using namespace std;int n;struct Node{int val;int pos;}node[maxn];int a[maxn];void update(int pos,int x){    while(pos<=n){        a[pos]+=x,pos+=lowbit(pos);    }}int getsum(int pos){    int res=0;    while(pos>0){        res+=a[pos],pos-=lowbit(pos);    }    return res;}bool cmp(const Node& a, const Node& b){return a.val > b.val;}int main(){    int T;    scanf("%d",&T);    int cont=0;while (T--){    scanf("%d",&n);    cont++;for (int i = 1; i <= n; ++i){scanf("%d", &node[i].val);node[i].pos = i;}sort(node + 1, node + n + 1, cmp);memset(a,0,sizeof(a));int ans=0;for(int i=1;i<=n;i++)        {            int t=getsum(node[i].pos);            if(node[i].pos-t!=node[i].val)                ans++,update(node[i].pos,1);        }printf("Case #%d: %d\n",cont, ans);}return 0;}

HDU 5115

裸区间DP

刚开始用双向链表写了个贪心策略的。看起来真优美啊，WA

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define inf 0x3f3f3f3f#define maxn 210typedef long long ll;using namespace std;int N,M,T;int V[maxn];int pre[maxn],next[maxn];struct P{    int pos,val,give;}p[maxn];bool cmp(const struct P &a,const struct P &b){    if(a.give==b.give)        return V[ pre[a.pos] ]+V[ next[a.pos] ] < V[ pre[b.pos]]+V[pre[b.pos]];    return a.give>b.give;}int main(){    scanf("%d",&T);    int cnt=0;while (T--){    scanf("%d",&N);ll ans=0;    for(int i=1;i<=N;i++)            scanf("%d",&p[i].val),p[i].pos=i;        for(int i=1;i<=N;i++)            scanf("%d",&p[i].give),V[i]=p[i].give;        V[0]=V[N+1]=0;        p[N+1].pos=N+1;        p[N+1].val=p[N+1].give=0;        p[0].pos=p[0].val=p[0].give=0;        for(int i=1;i<=N;i++)            next[i]=i+1,pre[i]=i-1;        sort(p+1,p+N+1,cmp);        ///memset(dp,0,sizeof(dp));        for(int i=1;i<=N;i++){           ans+=V[ pre[ p[i].pos ] ] +V[ next[ p[i].pos ] ] +p[i].val;           next[ pre[ p[i].pos ] ]=next[ p[i].pos ];           pre[ next[ p[i].pos ] ]=pre[ p[i].pos ];           sort(p+i+1,p+N+1,cmp);        }printf("Case #%d: %I64d\n",++cnt, ans);}return 0;}

AC:

#include <iostream>using namespace std;int main() {    int T;    cin >> T;    for (int cas = 1; cas <= T; cas++) {        int N;        cin >> N;        int a[202], b[202];        int dp[202][202];        for (int i = 1; i <= N; i++) {            cin >> a[i];        }        for (int i = 1; i <= N; i++) {            cin >> b[i];        }        b[0] = b[N + 1] = 0;        for (int i = 1; i <= N; i++) {            dp[i][i] = a[i] + b[i - 1] + b[i + 1];        }        for (int i = 1; i <= N ; i++) {            for (int j = 1; j + i <= N ; j++) {                dp[j][j + i] = -1;                for (int k = j; k <= j + i; k++) {                    int temp;                    if( k==j )                        temp = dp[k+1][j+i] + a[k] + b[j+i+1] + b[j-1];                    else if( k==i+j )                        temp = dp[j][k-1] + a[k] + b[j-1] + b[j+i+1];                    else                          temp = dp[j][k-1] + dp[k+1][j + i] + a[k] + b[j-1] + b[j + i + 1];                    if (dp[j][j + i] == -1 || temp < dp[j][j + i] ) {                        dp[j][j + i] = temp;                    }                }            }        }        cout << "Case #" << cas << ": " << dp[1][N] << endl;    }    return 0;}

HDU 5119

裸背包，放或不放的两种情况产生结果的转移。因为数字是小于1<<20，所以遍历到此即可。

注意2^40，要用Long long 因为这个卡了1次。

#include <cstdio>#include <cstring>#include <iostream>using namespace std;int N,M,T;long long dp[2][1<<20];int in[50];int main(){    scanf("%d",&T);    int cnt=0;    while(T--){        scanf("%d%d",&N,&M);        for(int i=1;i<=N;i++)            scanf("%d",in+i);        memset(dp,0,sizeof(dp));        dp[0][0]=1;        for(int i=1;i<=N;i++)            for(int j=0;j<(1<<20);j++)                    dp[i%2][j^in[i]]=dp[(i-1)%2][j]+dp[(i-1)%2][j^in[i]];        long long ans=0;        for(int i=M;i<(1<<20);i++)            ans+=dp[N%2][i];        printf("Case #%d: %I64d\n",++cnt,ans);    }}

HDU 5120

计算几何模板题，同学备好模板哦

#include<stdio.h>#include<math.h>const double eps = 1e-8;const double PI = acos(-1.0);double min(double a,double b) {    return a<b?a:b;}struct Point{double x,y;Point(){}Point(double _x,double _y){x = _x;y = _y;}Point operator -(const Point &b)const{return Point(x - b.x,y - b.y);}double operator *(const Point &b)const{return x*b.x + y*b.y;}};double dist(Point a,Point b){return sqrt((a-b)*(a-b));}double Area_of_overlap(Point c1,double r1,Point c2,double r2){double d = dist(c1,c2);if(r1 + r2 < d + eps)return 0;if(d < fabs(r1 - r2) + eps){double r = min(r1,r2);return PI*r*r;}double x = (d*d + r1*r1 - r2*r2)/(2*d);double t1 = acos(x / r1);double t2 = acos((d - x)/r2);return r1*r1*t1 + r2*r2*t2 - d*r1*sin(t1);}int main() {    #ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);    #endif    int T;    Point a,b;    double r,R;    double s;    scanf("%d",&T);    int cas=1;    while(T--) {        scanf("%lf%lf",&r,&R);        scanf("%lf%lf",&a.x,&a.y);        scanf("%lf%lf",&b.x,&b.y);        s=Area_of_overlap(a,R,b,R);        s-=Area_of_overlap(a,R,b,r);        s-=Area_of_overlap(a,r,b,R);        s+=Area_of_overlap(a,r,b,r);        printf("Case #%d: %.6lf\n",cas++,s);    }    return 0;}