哈理工软件学院"兆方美迪"杯第六届程序设计大赛【高年级组】--决赛 题解

比赛链接：http://acm-software.hrbust.edu.cn/contest.php?cid=1082

A.好SB啊真是，还以为lis…数有多少个数不一样。

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long LL;
 const int maxn = ;
 int ret;
 int n;
 set<int> s;
 int main() {
     // freopen("in", "r", stdin);
     int T;
     scanf("%d", &T);
     while(T--) {
         scanf("%d", &n);
         int x;
         s.clear();
         ret = ;
         for(int i = ; i < n; i++) {
             scanf("%d", &x);
             if(s.find(x) == s.end()) {
                 ret++;
                 s.insert(x);
             }
         }
         printf("%d\n", ret);
     }
     return ;
 }

B.顺着求和就行。

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long LL;
 const int maxn = ;
 LL ret;
 char s[maxn];

 int main() {
     // freopen("in", "r", stdin);
     int T;
     scanf("%d", &T);
     while(T--) {
         scanf("%s", s);
         ret = ;
         for(int i = ; s[i]; i++) {
             ret += s[i] - 'A' + ;
         }
         cout << ret << endl;
     }
     return ;
 }

C.折半1000-9999，然后对称过来。注意前导零。

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long LL;

 int main() {
     // freopen("in", "r", stdin);
     for(LL i = ; i <= ; i++) {
         printf("%d", i);
         int y = ;
         int t = i;
         int cnt = ;
         while(cnt--) {
             printf("%d", t % );
             t /= ;
         }
         printf("\n");
     }
     return ;
 }

D.并查集找到所有跟1能接触的人，做01背包。

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long LL;
 typedef struct P {
     int a, b, id;
 }P;

 const int maxn = ;
 int n, m, c;
 P p[maxn];
 int pre[maxn];
 int dp[maxn];
 int id[maxn];

 int find(int x) {
     return pre[x] == x ? x : pre[x] = find(pre[x]);
 }

 void unite(int x, int y) {
     x = find(x); y = find(y);
     if(x != y) pre[x] = y;
 }

 int main() {
     // freopen("in", "r", stdin);
     int T, u, v;
     scanf("%d", &T);
     while(T--) {
         scanf("%d%d%d",&n,&m,&c);
         memset(dp, , sizeof(dp));
         for(int i = ; i <= n; i++) pre[i] = i;
         for(int i = ; i <= n; i++) {
             scanf("%d %d", &p[i].a, &p[i].b);
             p[i].id = i;
         }
         for(int i = ; i < m; i++) {
             scanf("%d %d", &u, &v);
             unite(u, v);
         }
         int rt = find();
         for(int i = ; i <= n; i++) {
             if(rt != find(i)) continue;
             for(int j = c; j >= p[i].a; j--) {
                 dp[j] = max(dp[j], dp[j-p[i].a]+p[i].b);
             }
         }
         printf("%d\n", dp[c]);
     }
     return ;
 }

E.防ak不会…

F.双指针扫，卡住LOVE都出现的最短，然后向后扩展数一共有多少个。

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long LL;
 const int maxn = ;
 char s[maxn];
 int vis[];
 int n;
 int ret;
 int l, o, v, e;

 bool ok() {
     return l&&o&&v&&e;
 }

 void update(int i) {
     if(s[i]=='L')l++;
     if(s[i]=='O')o++;
     if(s[i]=='V')v++;
     if(s[i]=='E')e++;
 }

 int main() {
     // freopen("in", "r", stdin);
     int T;
     scanf("%d", &T);
     while(T--) {
         memset(s, , sizeof(s));
         scanf("%s", s+);
         n = strlen(s+);
         ret = ;
         int lo, hi;
         for(lo = ; lo <= n; lo++) {
             // memset(vis, 0, sizeof(vis));
             l=o=v=e=;
             for(hi = lo; hi <= n; hi++) {
                 // vis[s[hi]]++;
                 update(hi);
                 if(ok()) break;
             }
             if(hi <= n && ok()) ret = ret + n - hi + ;
         }
         printf("%d\n", ret);
     }
     return ;
 }

G.BFS，实现得有点SB。。。先特判第一步人能走到哪里。然后更新，在循环里先更新火的状态，这个时候在对内同一层的所有的人的状态面对的地图都是一样的。就可以用一个队列来做了。

 #include <bits/stdc++.h>
 using namespace std;

 typedef pair<int, int> pii;
 typedef struct P {
     int x, y, s;
     P() {}
     P(int x, int y, int s) : x(x), y(y), s(s) {}
 }P;
 const int inf = 0x7f7f7f;
 const int maxn = ;
 const int fdx[] = {-,-,-,,,,,};
 const int fdy[] = {-,,,-,,-,,};
 const int pdx[] = {-,,,};
 const int pdy[] = {,,-,};
 int n, m, ret;
 int sx, sy, ex, ey, fx, fy;
 int fcnt, pcnt;
 bool vis[maxn][maxn];
 char G[maxn][maxn];
 queue<P> q;

 bool ok(int x, int y) {
     return x >=  && x < n && y >=  && y < m;
 }

 bool nigero(int x, int y) {
     return x == ex && y == ey;
 }

 void bfs() {
     while(!q.empty()) q.pop();
     q.push(P(fx, fy, )); vis[fx][fy] = ; fcnt = ;
     int tmp = ;
     for(int i = ; i < ; i++) {
         int xx = sx + pdx[i];
         int yy = sy + pdy[i];
         if(nigero(xx, yy)) {
             ret = ;
             return;
         }
         if(ok(xx, yy) && G[xx][yy] != '*' && G[xx][yy] != '#' && !vis[xx][yy]) {
             vis[xx][yy] = ;
             q.push(P(xx, yy, )); tmp++;
         }
     }
     pcnt = tmp; tmp = ;
     while(!q.empty()) {
         for(int k = ; k < fcnt; k++) {
             P f = q.front(); q.pop();
             for(int i = ; i < ; i++) {
                 int xx = f.x + fdx[i];
                 int yy = f.y + fdy[i];
                 if(ok(xx, yy) && G[xx][yy] != '*') {
                     G[xx][yy] = '*';
                     q.push(P(xx, yy, f.s+)); tmp++;
                 }
             }
         }
         fcnt = tmp; tmp = ;
         for(int k = ; k < pcnt; k++) {
             P p = q.front(); q.pop();
             if(nigero(p.x, p.y)) {
                 ret = p.s;
                 return;
             }
             if(G[p.x][p.y] == '*') continue;
             for(int i = ; i < ; i++) {
                 int xx = p.x + pdx[i];
                 int yy = p.y + pdy[i];
                 if(ok(xx, yy) && G[xx][yy] != '*' && G[xx][yy] != '#' && !vis[xx][yy]) {
                     vis[xx][yy] = ;
                     q.push(P(xx, yy, p.s+)); tmp++;
                 }
             }
         }
         pcnt = tmp; tmp = ;
     }
 }

 int main() {
     // freopen("in", "r", stdin);
     int T;
     scanf("%d", &T);
     while(T--) {
         scanf("%d%d",&n,&m);
         ret = inf;
         for(int i = ; i < n; i++) scanf("%s", G[i]);
         memset(vis, , sizeof(vis));
         for(int i = ; i < n; i++) {
             for(int j = ; j < m; j++) {
                 if(G[i][j] == 'S') {
                     sx = i; sy = j;
                 } else if(G[i][j] == 'E') {
                     ex = i; ey = j;
                 } else if(G[i][j] == '*') {
                     fx = i; fy = j;
                 }
             }
         }
         bfs();
         if(ret == inf) puts("T_T");
         else printf("%d\n", ret);
     }
     return ;
 }

H.递推发现是个斐波那契，矩阵加速一下就行了。

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long LL;
 const LL mod = ;
 const int maxn = ;
 LL n;

 typedef struct Matrix {
     LL m[maxn][maxn];
     int r;
     int c;
     Matrix(){
         r = c = ;
         memset(m, , sizeof(m));
     }
 } Matrix;

 Matrix mul(Matrix m1, Matrix m2) {
     Matrix ans = Matrix();
     ans.r = m1.r;
     ans.c = m2.c;
     for(int i = ; i <= m1.r; i++) {
         for(int j = ; j <= m2.r; j++) {
                for(int k = ; k <= m2.c; k++) {
                 if(m2.m[j][k] == ) continue;
                 ans.m[i][k] = ((ans.m[i][k] + m1.m[i][j] * m2.m[j][k] % mod) % mod) % mod;
             }
         }
     }
     return ans;
 }

 Matrix quickmul(Matrix m, LL n) {
     Matrix ans = Matrix();
     for(int i = ; i <= m.r; i++) {
         ans.m[i][i]  = ;
     }
     ans.r = m.r;
     ans.c = m.c;
     while(n) {
         if(n & ) {
             ans = mul(m, ans);
         }
         m = mul(m, m);
         n >>= ;
     }
     return ans;
 }

 int main() {
     // freopen("in", "r", stdin);
     int T;
     scanf("%d", &T);
     while(T--) {
         scanf("%lld", &n);
         Matrix p, q;
         p.r = p.c = ;
         p.m[][] = ; p.m[][] = ;
         p.m[][] = ; p.m[][] = ;
         q.r = ; q.c = ;
         if(n <= ) {
             printf("%lld\n", n);
             continue;
         }
         q = quickmul(p, n-);
         printf("%lld\n", (q.m[][] + q.m[][]) % mod);
     }
     return ;
 }

I.求n次最短路，每次求最短路的时候是以那个点为枢纽。给所有的最短路排序，然后从大到小找两个点，把两条最长的最短路加起来更新答案。

 #include <bits/stdc++.h>
 using namespace std;

 typedef pair<int, int> pii;
 const int inf = 0x7f7f7f7;
 const int maxn = ;
 vector<pii> G[maxn];
 bool vis[maxn];
 queue<int> q;
 int d[maxn];
 int n, m, ret;

 void spfa(int s) {
     for(int i = ; i <= n; i++) d[i] = inf;
     memset(vis, , sizeof(vis));
     while(!q.empty()) q.pop();
     d[s] = ; q.push(s);
     while(!q.empty()) {
         int u = q.front(); q.pop();
         vis[u] = ;
         for(int i = ; i < G[u].size(); i++) {
             int& v = G[u][i].first;
             int& w = G[u][i].second;
             if(d[v] > d[u] + w) {
                 d[v] = d[u] + w;
                 if(vis[v]) continue;
                 vis[v] = ; q.push(v);
             }
         }
     }
     sort(d+, d+n+);
     int cnt = , x = ;
     for(int i = n; i >= ; i--) {
         if(cnt >= ) break;
         if(d[i] == ) continue;
         if(d[i] != inf) {
             x += d[i];
             cnt++;
         }
     }
     if(cnt == ) ret = max(ret, x);
 }

 int main() {
     // freopen("in", "r", stdin);
     int u, v, w, T;
     scanf("%d", &T);
     while(T--) {
         ret = -;
         scanf("%d%d",&n,&m);
         for(int i = ; i <= n; i++) G[i].clear();
         for(int i = ; i < m; i++) {
             scanf("%d%d%d", &u,&v,&w);
             G[u].push_back(pii(v, w));
             G[v].push_back(pii(u, w));
         }
         for(int i = ; i <= n; i++) spfa(i);
         printf("%d\n", ret);
     }
     return ;
 }