【URAL水题】集中题解区

列出了难度在150以上的水题，150以下的就不列出了

http://acm.timus.ru/problem.aspx?space=1&num=1346

这道是动规题，我模拟做的，划分区间，找出最少的单调区间数量之和。

我想到了一个方法，但是不能证明。但是可以AC。

就是先求出该离散函数的所有单调区间之和，如果有连线后，有两个连续的拐点，则可以把这两点分开，分别包含于两个区间，对答案会减一。

但是如果说再有一个拐点紧接在后面的话就不能继续这样操作了。

/*感觉有规律，但是没法证明。。*\\* 可能不完美。对付URAL的测试足够了*/ #include <cstdio>long a;long b;long num[100010];long n = 0;int main(){scanf("%ld%ld",&a,&b);b -= a-1;a = 1;long last = 0;for (long i=1;i<b+1;i++){scanf("%ld",num+i);if (num[i]!=last){last = num[i];num[++n] = num[i];}}long cnt = 1;long inc = num[2]>num[1]?1:-1;long continus = 0;for (long i=2;i<n+1;i++){long tmp = num[i]>num[i-1]?1:-1;if (tmp != inc){inc = -inc;cnt ++;continus++;}else{continus = 0;}/**\连续两个拐点，则从中间断开如果连续三个拐点，只能断开其中之一 \**/if (continus == 2){cnt --;continus = 0;}}printf("%ld",cnt);return 0;}

http://acm.timus.ru/problem.aspx?space=1&num=1167

简单的动规，线性的，也许可以单调队列优化，但是懒得了。用前缀和预处理出黑色马和白色马的数量。

注意宏定义的格式宏定义中一定要给每个参数加括号在最外层加括号，否则很容易出错

#include <cstdio>#include <iostream>long n;long K;long col[510];long sum[510];long f[510][510];//long BLACK(long a,long b){return (sum[b]-sum[a-1]);}//long WHITE(long a,long b){return ((b-a+1)-(sum[b]-sum[a-1]));}#define BLACK(a,b) (sum[b]-sum[a-1])#define WHITE(a,b) (((b)-(a)+1)-(sum[(b)]-sum[(a)-1]))/**\宏定义中一定要给每个参数加括号在最外层加括号 \**/#define MIN(a,b) (a<b?a:b)int main(){scanf("%ld%ld",&n,&K);for (long i=1;i<n+1;i++){scanf("%ld",col+i);sum[i] = sum[i-1] + col[i];}for (long i=0;i<n+1;i++)for (long j=0;j<K+1;j++)f[i][j] = 0x7f7f7f7f;f[0][0] = 0;for (long i=1;i<n+1;i++){#ifdef Debugstd::cerr << i << std::endl; #endiffor (long j=1;j<K+1&&j<i+1;j++){f[i][j] = 0x7f7f7f7f;for (long k=0;k<i;k++){f[i][j] = MIN(f[i][j],f[k][j-1]+BLACK(k+1,i)*WHITE(k+1,i));}//f[i][j] = MIN(f[i][j],f[i-1][j]);}}printf("%ld",f[n][K]);return 0;}

http://acm.timus.ru/problem.aspx?space=1&num=1078

就是求最多有几个严格上一个包含下一个的区间。

很简单，先排一次序，第一关键字按l的升序，第二关键字按r的降序。并且把0号点区间设为(-inf,inf)

然后就是一个二维的动规。

#include <cstdio>#include <algorithm>long n;const long inf = 0x7f7f7f7f;long g[510];struct node{long l;long r;long i;bool operator<(const node& n2)const {if (l < n2.l) return true;if (l > n2.l) return false;if (r < n2.r) return false;return true;}};node line[510];long f[510];long output[510];long top = 0;#define MAX(a,b) (a>b?a:b)int main(){scanf("%ld",&n);for (long i=1;i<n+1;i++){scanf("%ld%ld",&line[i].l,&line[i].r);line[i].i = i;}std::sort(line+1,line+n+1);line[0].l = -inf;line[0].r = inf;long ans = 0;long des = 0;for (long i=1;i<n+1;i++){for (long j=0;j<i;j++){if (line[j].r > line[i].r && line[j].l < line[i].l){if (f[j] + 1 > f[i]){f[i] = f[j] + 1;g[i] = j;}}}if (f[i] > ans){ans = f[i];des = i;}}printf("%ld\n",ans);long i = des;while (i){output[++top] = line[i].i;i = g[i];}for (long i=1;i<top+1;i++){printf("%ld ",output[i]);}return 0;}

http://acm.timus.ru/problem.aspx?space=1&num=1157

年轻的瓦工。规模小。朴素方法就过了。

#include <cstdio>#include <cmath>long m;long n;long k;long cnt = 0;long ans(long x){long t = 0;//for (long i=1;i<=ceil(sqrt(double(x)));i++)for (long i=1;i<=long(sqrt(double(x))+0.5);i++){if (x % i == 0) t++;}return t;}int main(){scanf("%ld%ld%ld",&m,&n,&k);for (long i=k+1;i<10001;i++){if (ans(i)==n&&ans(i-k)==m){printf("%ld\n",i);return 0;}} printf("0\n");return 0;}

http://acm.timus.ru/problem.aspx?space=1&num=1029

貌似是那个什么数字三角形的变形。

逐层推。由于拓扑序不明显，有点乱，又是从左到右，又是从右到左，因此每一层用了两个临时数组来维护两个转移，然后再用两个数组更新F数组

#include <cstdio>typedef long long ll;long m;long n;ll f1[510];ll f2[510];long g1[510];long g2[510];long g[110][510];ll f[110][510];long map[110][510];long op[50010];long top = 0;int main(){scanf("%ld%ld",&n,&m);for (long i=1;i<n+1;i++){for (long j=1;j<m+1;j++){scanf("%ld",map[i]+j);}}for (long i=1;i<n+1;i++){f1[0] = f1[m+1] =f2[0] = f2[m+1] = 0x3f7f7f7f7f7f7f7fll;for (long j=1;j<m+1;j++)if (f[i-1][j]+(ll)map[i][j] < f1[j-1]+(ll)map[i][j]){f1[j] = f[i-1][j]+(ll)map[i][j];g1[j] = j;}else{f1[j] = f1[j-1]+(ll)map[i][j];g1[j] = j-1;}for (long j=m;j>0;j--)if (f[i-1][j]+(ll)map[i][j] < f2[j+1]+(ll)map[i][j]){f2[j] = f[i-1][j]+(ll)map[i][j];g2[j] = j;}else{f2[j] = f2[j+1]+(ll)map[i][j];g2[j] = j+1;}for (long j=1;j<m+1;j++){if (f1[j] < f2[j]){f[i][j] = f1[j];g[i][j] = g1[j];}else{f[i][j] = f2[j];g[i][j] = g2[j];}}}ll ans = 0x7f7f7f7f7f7f7f7fll;long des = 0;for (long j=1;j<m+1;j++){if (ans > f[n][j]){ans = f[n][j];des = j;}}long i = n;long j = des;while (g[i][j]){if (j == g[i][j]){op[++top] = j;i --;}else{op[++top] = j;j = g[i][j];}}for (long i=top;i>0;i--){printf("%ld ",op[i]);}//printf("%ld",ans);return 0;}

http://acm.timus.ru/problem.aspx?space=1&num=1002

本来是字符串匹配+DP。但是C++优势，字符串匹配就直接搞定了。转移就很简单了。

#include <iostream>#include <algorithm>#include <cstdio>#include <string>using std::min;using std::string;using std::cout;using std::cin;string PN;string wrds[50010];string wrds2[50010];char chrctr[26] = {'2','2','2','3','3','3','4','4','1','1','5','5','6','6','0','7','0','7','7','8','8','8','9','9','9','0'};long length[50010];long f[50010];long g[50010];long pnlen;long n;void dp(){    for (long i=0;i<pnlen;i++)    {        g[i] = 0;        f[i] = 0x7f7f7f7f;    }    for (long i=1;i<n+1;i++)    {        if (PN.compare(0,length[i],wrds[i],0,length[i])==0)        {            f[length[i]-1] = 1;            g[length[i]-1] = i;        }    }    for (long i=1;i<pnlen;i++)    {        for (long j=1;j<n+1;j++)        {            if (i-length[j]>=0 && PN.compare(i-length[j]+1,length[j],wrds[j],0,length[j])==0)            {                if (f[i-length[j]]+1 < f[i])                {                    f[i] = f[i-length[j]]+1;                    g[i] = j;                }            }        }    }}void ou(long l){    if (g[l] == 0)return;    ou(l-length[g[l]]);    cout << wrds2[g[l]] << " ";}void translate(long l){    for (long i=0;i<length[l];i++)    {        wrds[l][i] = chrctr[wrds[l][i]-'a'];    }}int main(){    freopen("phone.in","r",stdin);    freopen("phone.out","w",stdout);    while (1)    {        cin >> PN;        if (PN == "-1") break;        pnlen = PN.length();        cin >> n;        for (long i=1;i<n+1;++i)        {            cin >> wrds[i];            wrds2[i] = wrds[i];            length[i] = wrds[i].length();            translate(i);        }        dp();        if (f[pnlen-1]<0x7f7f7f7f) ou(pnlen-1);        else cout << "No solution.\n";        cout << "\n";    }}