Zhejiang Provincial Programming Contest 2007 部分题解

AAttack of Panda Virus    

优先队列 +　bfs

可以根据def 和 到可能到达此处的病毒等级为元素进行排列

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <vector>#include <map>using namespace std;const int LMT = 250003;int have[LMT], vis[502][502],cnt;int gra[502][502];map<int, int> lev_mp;struct __node{    int d, lev, posi, posj, mlev;    __node(){}    __node(int a, int b, int c, int i, int j)    : d(a), lev(b), mlev(c), posi(i), posj(j){}    bool operator > (const __node &y)    {        if(d != y.d) return d > y.d;        return lev > y.lev;    }};struct cmp{    bool operator ()(__node a, __node b)    {        return a > b;    }};priority_queue<__node, vector<__node>, cmp> que;void init(void){    cnt = 0;    while(!que.empty())que.pop();    lev_mp.clear();    memset(vis, 0, sizeof(vis));    memset(have, 0, sizeof(have));}int main(void){    int n, m, alf, temi, temj;    __node from, to;    while(~scanf("%d%d", &n, &m))    {        init();        alf = 0;        for(int i = 0; i < n; ++i)          for(int j = 0; j < m; ++j)          {             scanf("%d", &gra[i][j]);             if(gra[i][j] > 0)             {               if(lev_mp.find(gra[i][j]) == lev_mp.end())               {                   lev_mp.insert(make_pair(gra[i][j], cnt));                   que.push(__node(0, gra[i][j], cnt, i, j));                   ++cnt;               }             }          }          while(!que.empty())          {              from = que.top();              temi = que.top().posi;              temj = que.top().posj;              que.pop();              if(vis[temi][temj])continue;              vis[temi][temj] = 1;              ++have[from.mlev];              ++alf;              if(alf >= n * m) break;              if(temi > 0 && 0 == vis[temi - 1][temj])              {                  if(gra[temi - 1][temj] < 0)                  que.push                  (                  __node(-gra[temi - 1][temj],                    from.lev, from.mlev,temi - 1, temj));              }              if(temj >0 && 0 == vis[temi][temj - 1])              {                  if(gra[temi][temj - 1] < 0)              que.push              (              __node(-gra[temi][temj - 1],               from.lev, from.mlev, temi, temj - 1));              }               if(temi < n - 1 && 0 == vis[temi + 1][temj])               {                   if(gra[temi + 1][temj] < 0)               que.push               (               __node(-gra[temi + 1][temj],from.lev, from.mlev, temi + 1, temj));               }               if(temj < m - 1 && 0 == vis[temi][temj + 1])               {                   if(gra[temi][temj + 1] < 0)               que.push               (               __node(-gra[temi][temj + 1],from.lev, from.mlev, temi, temj + 1));               }          }          int q, ql;          scanf("%d", &q);          while(q--)          {              scanf("%d", &ql);              if(lev_mp.find(ql) != lev_mp.end())              printf("%d\n", have[lev_mp[ql]]);              else printf("0\n");          }    }    return 0;}

B 水

C Code Formatter

字符串处理，WA了好多次， 注意终止条件是一行“只含有”两个‘#’

#include <cstdio>#include <cstring>const int LMT = 102;char sec[LMT];int main(void){int i,tab, tail, T, len;scanf("%d", &T);getchar();while(T--){tab = 0; tail = 0;while(gets(sec)){len = strlen(sec);if(len == 2 && sec[0] == '#' && sec[1] == '#') break;for(i = len - 1; i >= 0; --i)if (sec[i] == ' ')++tail;else if(sec[i] == '\t')tail +=4;else break;            for(i = 0; i < len; ++i)if(sec[i] == '\t')++tab;}printf("%d tab(s) replaced\n%d trailing space(s) removed", tab, tail);if (T) printf("\n"); }return 0;}

E Deck of Cards 

水的动态规划， 当前的状态是选择哪一个洞， 注意当一个洞放上F或上面的牌点数总和等于21时，有额外积分，并且可以把上面的牌都拿掉

#include <cstdio>#include <cstring>#define max(a, b) ((a) > (b) ? (a) : (b))#define min(a, b) ((a) < (b) ? (a) : (b))const int LMT = 23;int dp[103][LMT][LMT][LMT];int vn[103];int main(void){int val[100], n, i, x, y, z, tem, ans;char ch;val['A'] = 1; val['6'] = 6;val['2'] = 2; val['7'] = 7;val['3'] = 3; val['8'] = 8;val['4'] = 4; val['9'] = 9;val['5'] = 5; val['T'] = 10;val['J'] = 10; val['Q'] = 10;val['K'] = 10; val['F'] = 22;while(~scanf("%d", &n) && n){memset(dp, -1 , sizeof(dp));ans = -1;dp[0][0][0][0] = 0;for(i = 1; i <= n; ++i){ch = getchar();while(ch == ' ' || ch == '\n')ch = getchar();vn[i] = val[ch];}for(i = 1; i <= n; ++i){for(x = 0; x <= 22; ++x)for(y = 0; y <= 22; ++y)for(z = 0; z <= 22; ++z)if(dp[i - 1][x][y][z] != -1){if(x < 22){tem = x + vn[i];if(tem > 22) tem =22;if(tem == 21 || vn[i] == 22) tem =0;if(tem)   dp[i][tem][y][z] = max(dp[i][tem][y][z], dp[i - 1][x][y][z] + 50);else dp[i][tem][y][z] = max(dp[i][tem][y][z], dp[i - 1][x][y][z] + 150);if(i == n || tem >=22 || y >= 22 || z >= 22)ans = max(ans, dp[i][tem][y][z]);}if(y < 22){tem = y + vn[i];if(tem > 22) tem =22;if(tem == 21 || vn[i] == 22) tem =0;if(tem)  dp[i][x][tem][z] = max(dp[i][x][tem][z], dp[i - 1][x][y][z] + 50);else dp[i][x][tem][z] = max(dp[i][x][tem][z], dp[i - 1][x][y][z] + 250);if(i == n || x >=22 || tem >= 22 || z >= 22)ans = max(ans, dp[i][x][tem][z]);}if(z < 22){tem = z + vn[i];if(tem > 22) tem =22;if(tem == 21 || vn[i] == 22) tem =0;if(tem)dp[i][x][y][tem] = max(dp[i][x][y][tem], dp[i - 1][x][y][z] + 50);else dp[i][x][y][tem] = max(dp[i][x][y][tem], dp[i - 1][x][y][z] + 350);if(i == n || x >=22 || y >= 22 || tem >= 22)ans = max(ans, dp[i][x][y][tem]);}}}printf("%d\n", ans);}return 0;}

E 矩阵乘法，小心点就好了

F 计算几何， 不会啊...

G google map 

因为远点计算出错，卡了...

#include<cstdio>#include<cmath>#define eps 1e-9const long double pi = acos(-1);inline double lab_mid(double l, double r){double tem = atan(   sqrt(tan(pi / 4 + (l * pi / 180) / 2) *              tan(pi / 4 + (r * pi / 180) / 2)));return (tem - pi/4) * 2 * 180 / pi;}int main(void){double loq, laq, lab, lob;double lau, lad, lol, lor, temlo, temla;char tag;int L;while(~scanf("%lf%lf%d", &loq, &laq, &L)){    temla = temlo = lob = lab = 0.0;lol = -180.0; lor = 180.0;lau = 85; lad = -85;printf("t");for(int i = 1; i <=L; ++i){if(loq > lob&& laq > lab){tag = 'r';lob = (lob + lor)*0.5;lab = lab_mid(lab, lau);lol = temlo; lor = lor;lad = temla; lau = lau;}else if(loq > lob && laq < lab){tag = 's';lob = (lob + lor)*0.5;lab = lab_mid(lad, lab);lol = temlo; lor = lor;lad = lad; lau = temla;}else if(loq < lob && laq > lab){tag = 'q';lob = (lob + lol) * 0.5;lab = lab_mid(lab, lau);lol = lol; lor = temlo;lad = temla; lau = lau;}else if(loq < lob && laq < lab){tag = 't';lob = (lob + lol) * 0.5;lab = lab_mid(lad, lab);lol = lol; lor = temlo;lad = lad; lau = temla;}printf("%c", tag);temlo = lob;temla = lab;}printf("\n");}return 0;}