CSU Monthly 2013 Oct 中南大学ACM月赛

题目很好，比较活跃，有难度，同时对基础要求又比较好，总体感觉这题目做起来挺不错的，没做出的去补一下

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1318

数学题：

#include<iostream>#include<cstdio>#include<list>#include<algorithm>#include<cstring>#include<string>#include<queue>#include<stack>#include<map>#include<vector>#include<cmath>#include<memory.h>#include<set>#define ll long long#define eps 1e-8#define inf 0xfffffff//const ll INF = 1ll<<61;using namespace std;//vector<pair<int,int> > G;//typedef pair<int,int > P;//vector<pair<int,int> > ::iterator iter;////map<ll,int >mp;//map<ll,int >::iterator p;int main() {ll n;while(scanf("%lld",&n) == 1) {ll ans = 1;ll num = 1;while((num*2) - 1 < n) {ans++;num *= 2;}printf("%lld\n",ans);}return EXIT_SUCCESS;}

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1321

spfa，记录最短路径的同时记录妹子的数量

#include<iostream>#include<cstdio>#include<list>#include<algorithm>#include<cstring>#include<string>#include<queue>#include<stack>#include<map>#include<vector>#include<cmath>#include<memory.h>#include<set>#define ll long long#define eps 1e-8#define inf 0xfffffff//const ll INF = 1ll<<61;using namespace std;//vector<pair<int,int> > G;//typedef pair<int,int > P;//vector<pair<int,int> > ::iterator iter;////map<ll,int >mp;//map<ll,int >::iterator p;int n,m;bool vis[10000 + 5];int dis[10000 + 5];int ans[10000 + 5];int num [10000 + 5];int head[10000 + 5];typedef struct Node {int from,to;int nex;int w;};Node edge[100000 + 5];int tot;void clear() {memset(num,0,sizeof(num));memset(edge,0,sizeof(edge));memset(head,-1,sizeof(head));tot = 0;}void add(int u,int v,int w) {edge[tot].from = u;edge[tot].to = v;edge[tot].nex = head[u];edge[tot].w = w;head[u] = tot++;}int spfa(int s,int e) {memset(vis,false,sizeof(vis));for(int i=0;i<=n;i++) dis[i] = inf;for(int i=1;i<=n;i++)ans[i] = inf;dis[s] = 0;ans[s] = num[s];vis[s] = true;queue<int> q;q.push(s);while(!q.empty()) {int u = q.front();q.pop();vis[u] = false;for(int i=head[u];i!=-1;i=edge[i].nex) {int v = edge[i].to;if(dis[u] + edge[i].w < dis[v]) {dis[v] = edge[i].w + dis[u];ans[v] = ans[u] + num[v];if(!vis[v]) {vis[v] = true;q.push(v);}}else if(dis[v] == dis[u] + edge[i].w) {if(ans[v] < ans[u] + num[v]) {ans[v] = ans[u] + num[v];if(!vis[v]) {vis[v] = true;q.push(v);}}}}}if(dis[n] == inf) return -1;elsereturn ans[n];}int main() {while(scanf("%d %d",&n,&m) == 2) {clear();for(int i=1;i<=n;i++)scanf("%d",&num[i]);for(int i=1;i<=m;i++) {int u,v,w;scanf("%d %d %d",&u,&v,&w);add(u,v,w);add(v,u,w);}int answer = spfa(1,n);printf("%d\n",answer);}return EXIT_SUCCESS;}

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1320

仔细写几个，不能遗漏，发现是卡特兰数，因为要取模 涉及到整除问题，所以要用扩展欧几里德求乘法逆元来避免这个问题

#include<iostream>#include<cstdio>#include<list>#include<algorithm>#include<cstring>#include<string>#include<queue>#include<stack>#include<map>#include<vector>#include<cmath>#include<memory.h>#include<set>#define ll long long#define eps 1e-8#define inf 0xfffffff//const ll INF = 1ll<<61;using namespace std;//vector<pair<int,int> > G;//typedef pair<int,int > P;//vector<pair<int,int> > ::iterator iter;////map<ll,int >mp;//map<ll,int >::iterator p;const ll MOD = 1000000007;ll dp[10000 + 5];ll n;ll exgcd(ll a, ll b, ll &x, ll &y)  {if(!b) {x = 1; y = 0;return a;}ll r = exgcd(b, a%b, y, x);y -= a/b*x;return r;}ll inv(ll a, ll m) //求逆元直接模版套上 {ll x,y,gcd = exgcd(a, m, x, y);if(x < 0)  x += m;return x;}void clear() {memset(dp,0,sizeof(dp));dp[1] = 1;dp[2] = 2;for(int i=3;i<10001;i++) {dp[i] = ((dp[i-1]*(4*i-2)%MOD) * ((inv(i+1,MOD) + MOD)%MOD))%MOD;}}int main() {clear();while(scanf("%lld",&n) == 1) {printf("%lld\n",dp[n]);}return EXIT_SUCCESS;}

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1326

符合分组背包，所以用分组背包做一下，注意分组要求

#include<iostream>#include<cstdio>#include<list>#include<algorithm>#include<cstring>#include<string>#include<queue>#include<stack>#include<map>#include<vector>#include<cmath>#include<memory.h>#include<set>#define ll long long#define eps 1e-8#define inf 0xfffffff//const ll INF = 1ll<<61;using namespace std;//vector<pair<int,int> > G;//typedef pair<int,int > P;//vector<pair<int,int> > ::iterator iter;////map<ll,int >mp;//map<ll,int >::iterator p;vector<int> G[1000 + 5];int n,V,q;int dp[1000 + 5];int p[1000 + 5],w[1000 + 5];int father[1000 + 5];void clear() {memset(dp,0,sizeof(dp));for(int i=0;i<1005;i++)father[i] = i;for(int i=0;i<1005;i++)G[i].clear();}int find(int x) {if(father[x] != x)return find(father[x]);return x;}void merge(int x,int y) {int dx = find(x);int dy = find(y);if(dx!=dy) {father[dx] = dy;}}int main() {while(scanf("%d %d %d",&n,&V,&q) == 3) {clear();for(int i=0;i<n;i++)scanf("%d %d",&p[i],&w[i]);for(int i=0;i<q;i++) {int tx,ty;scanf("%d %d",&tx,&ty);tx--;ty--;merge(tx,ty);}for (int i=0;i<n;i++) {int x = find(i);G[x].push_back(i);}for (int i=0;i<n;i++) {if (G[i].size() != 1)continue;for (int j=V;j>=w[i];j--) if (dp[j - w[i]] + p[i] > dp[j])dp[j] = dp[j - w[i]] + p[i];}for (int i=0;i<n;i++) {if (G[i].size() == 1) continue;for (int k=V;k>=0;k--) {for (int j=0;j<G[i].size();j++) {if (w[G[i][j]] > k) continue;if (dp[k - w[G[i][j]]] + p[G[i][j]] > dp[k])dp[k] = dp[k - w[G[i][j]]] + p[G[i][j]];}}}printf("%d\n",dp[V]);}return EXIT_SUCCESS;}