期望得分：100+100+100=300
实际得分：0+100+90=190
T1 superman
二分给每条边加多少，判断是否存在负环
#include#include #include #include #define N 501#define M 4951using namespace std;int n,tot,front[N],to[M],nxt[M],val[M],val_[M];int front_[N],to_[M],nxt_[M];int cnt[N],dis[N];bool vis[N],ok[N];void add(int u,int v,int w){ to[++tot]=v; nxt[tot]=front[u]; front[u]=tot; val_[tot]=w; to_[tot]=u; nxt_[tot]=front_[v]; front_[v]=tot;}bool spfa(){ memset(cnt,0,sizeof(cnt)); memset(dis,63,sizeof(dis)); memset(vis,false,sizeof(vis)); queue q; vis[1]=true; q.push(1); dis[1]=0; cnt[1]++; int now; while(!q.empty()) { now=q.front(); q.pop(); vis[now]=false; for(int i=front[now];i;i=nxt[i]) { if(!ok[to[i]]) continue; if(dis[to[i]]>dis[now]+val[i]) { dis[to[i]]=dis[now]+val[i]; if(!vis[to[i]]) { vis[to[i]]=true; q.push(to[i]); cnt[to[i]]++; if(cnt[to[i]]>n) return false; } } } } return dis[n]>=0;}bool check(int mid){ for(int i=1;i<=tot;i++) val[i]=val_[i]+mid; return spfa();}void solve(int tmp){ for(int i=1;i<=tot;i++) val[i]=val_[i]+tmp; spfa(); printf("%d\n",dis[n]);}void dfs(int x){ ok[x]=true; for(int i=front[x];i;i=nxt[i]) if(!ok[to[i]]) dfs(to[i]);}void dfs2(int x){ ok[x]=true; for(int i=front_[x];i;i=nxt_[i]) if(!ok[to_[i]]) dfs2(to_[i]);}int main(){ freopen("superman.in","r",stdin); freopen("superman.out","w",stdout); int T,m,u,v,w; scanf("%d",&T); while(T--) { tot=0; memset(front,0,sizeof(front)); memset(front_,0,sizeof(front_)); scanf("%d%d",&n,&m); for(int i=1;i<=m;i++) { scanf("%d%d%d",&u,&v,&w); add(u,v,w); } memset(ok,false,sizeof(ok)); dfs(1); if(!ok[n]) { printf("-1\n"); continue; } memset(ok,false,sizeof(ok)); dfs2(n); int l=-100000,r=100000,mid,tmp; while(l<=r) { mid=(l+r)/2; if(check(mid)) tmp=mid,r=mid-1; else l=mid+1; } solve(tmp); } }
T2 洛谷 P2647 最大收益
如果一共选m个，第i个选了第j个物品，那对答案的贡献为wj-（m-i）*rj
所以如果确定选哪m个，肯定是先选rj小的
去除m的影响，倒着枚举
AC代码
#include#include #define N 3001using namespace std;struct node{ int w,r;}e[N]; int wi[N],ri[N],dp[N][N];bool cmp(node p,node q){ return p.r>q.r;}int main(){ //freopen("market.in","r",stdin); //freopen("market.out","w",stdout); int n,ans=0; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d%d",&e[i].w,&e[i].r); sort(e+1,e+n+1,cmp); for(int i=1;i<=n;i++) for(int j=1;j<=i;j++) dp[i][j]=max(dp[i-1][j],dp[i-1][j-1]+e[i].w-e[i].r*(j-1)); for(int i=1;i<=n;i++) ans=max(ans,dp[n][i]); printf("%d",ans); }
暴力1
#include#include #define N 3001using namespace std;struct node{ int w,r;}e[N]; int wi[N],ri[N],dp[N][N];bool cmp(node p,node q){ return p.r
暴力2
#include#include #define N 3001using namespace std;struct node{ int w,r;}e[N]; int wi[N],ri[N];bool cmp(node p,node q){ return p.r
T3 洛谷 P2759 奇怪的函数
x>=logx 10^(n-1)
x>=[log10 10^(n-1)]/log 10 x
x>=(n-1)/log10 x
特判当x=1时，答案为1
#include#include using namespace std;typedef long long LL;int n;int main(){ //freopen("Lemon_Soda.in","r",stdin); //freopen("Lemon_Soda.out","w",stdout); scanf("%d",&n); LL l=1,r=1ll<<62,mid,ans; if(n==1) { printf("1"); return 0; } while(l<=r) { mid=l+r>>1; if((n-1)*1.0/log10(mid)<=mid) r=mid-1,ans=mid; else l=mid+1; } printf("%lld",ans);}