NOIP2016 Topic Integration

Today, I finally got the official data, and I had a full-on.

Day 1 T1 Toy

Deal with the positive and negative number of the minus-minus to take the mold to mess up just fine.

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype > #include <algorithm>using namespace Std;int read () {    int x = 0, f = 1; char c = GetChar ();    while (!isdigit (c)) {if (c = = '-') f =-1; c = GetChar ();}    while (IsDigit (c)) {x = x * + C-' 0 '; c = GetChar ();}    return x * f;} #define MAXN 100010int N, q;struct Toy {int tp; char s[15];} ts[maxn];int Main () {freopen ("toy.in", "R", stdin); Freopen ( "Toy.out", "w", stdout);    n = read (); Q = Read ();    for (int i = 0; i < n; i++) {        int t = read (); scanf ("%s", Ts[i]. S);        if (!t) ts[i].tp = 1;        else TS[I].TP =-1;    }    int p = 0;    while (q--) {        int a = read (), B = Read ();        if (a) a = 1; else A =-1;        p + = A * TS[P].TP * b;        p = (p% n + N)% n;    }        printf ("%s\n", Ts[p]. S);        return 0;}

Day 1 T2 Running

The

is inspired by the special points "chain" and "S constant 1" and "T constant 1", we found that each chain [Si, Ti] can be divided into [Si, ci] and [CI, Ti] Two, and then for a all observable chain  [ci, Ti] will all Wi minus depth is a Fixed value, for the [Si, Ci] portion of all Wi plus depth is a fixed value. So the tree chain split and then hit the marker statistics just fine.

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype    > #include <algorithm>using namespace Std;int read () {int x = 0, f = 1; char c = GetChar ();    while (!isdigit (c)) {if (c = = '-') f =-1; c = GetChar ();}    while (IsDigit (c)) {x = x * + C-' 0 '; c = GetChar ();} return x * f;} #define MAXN 300010#define maxm 600010#define maxs 12000010int N, Q, M, HEAD[MAXN], NEXT[MAXM], TO[MAXM], w[maxn];struct P    layer {int S, T, C; Player () {} player (int _, int __): s (_), T (__) {}} ps[maxn];void Addedge (int a, int b) {To[++m] = b; next[m] = head [A];    Head[a] = m;    Swap (A, b); TO[++M] = b; NEXT[M] = Head[a];    Head[a] = m; return;}    int FA[MAXN], DEP[MAXN], SIZ[MAXN], SON[MAXN], TOP[MAXN], POS[MAXN], PID[MAXN], clo;void build (int u) {siz[u] = 1;        for (int e = head[u]; e; e = next[e]) if (to[e]! = Fa[u]) {fa[to[e]] = u;        Dep[to[e]] = Dep[u] + 1;        Build (To[e]); Siz[u] + = Siz[to[E]];    if (Siz[son[u]] < Siz[to[e]]) son[u] = To[e]; } return;    void Gett (int u, int tp) {Top[u] = TP; Pid[++clo] = u; pos[u] = CLO;    if (Son[u]) Gett (Son[u], TP);    for (int e = head[u]; e; e = next[e]) if (to[e]! = Fa[u] && to[e]! = Son[u]) Gett (To[e], to[e]); return;}    int LCA (int a, int b) {int f1 = Top[a], F2 = top[b];        while (f1! = F2) {if (Dep[f1] < DEP[F2]) swap (F1, F2), swap (A, b); A = Fa[f1];    F1 = Top[a]; } return Dep[a] < Dep[b]? A:B;}    struct Info {int C, FIR[MAXN], Aft[maxs], VAL[MAXS];        void Clear () {c = 0;        memset (fir, 0, sizeof (FIR));    return;        } void Addinfo (int x, int v) {Val[++c] = v; aft[c] = fir[x]; fir[x] = c;    return;    }} Add, Del;int tot[maxn<<1], ans[maxn];void process (int x, int t, int a, int v) {int f = top[a]; while (f! = Top[t]) {Add.        Addinfo (Pos[f], V); Del.        Addinfo (Pos[a], V); A = Fa[f];    f = top[a]; } ADD.addinfo (Pos[t], V); Del.    Addinfo (Pos[a], V); return;}    int main () {freopen ("running.in", "R", stdin), Freopen ("Running.out", "w", stdout); n = read ();    Q = Read ();        for (int i = 1; i < n; i++) {int a = read (), B = Read ();    Addedge (A, b);    } for (int i = 1; I <= n; i++) W[i] = read ();        for (int i = 1; I <= Q; i++) {int s = read (), t = Read ();    Ps[i] = Player (s, t);    } build (1);        Gett (1, 1);    for (int i = 1; I <= n; i++) w[i] = Dep[i]; Add.clear (); Del.clear ();    memset (tot, 0, sizeof (TOT));        for (int i = 1; I <= Q; i++) {ps[i].c = LCA (PS[I].S, ps[i].t);    Process (I, PS[I].C, ps[i].t, Dep[ps[i].s]-dep[ps[i].c]-dep[ps[i].c]); } for (int i = 1, i <= N; i++) {for (int e = add.fir[i]; e = add.aft[e]) {int v = add.val[e] +            N        tot[v]++;        } int u = pid[i];        Ans[u] + = Tot[w[u]+n]; for (int e = del.fir[i], e; e = del.aft[e]) {int v = Del.val[e] + N;        tot[v]--;    }} for (int i = 1; I <= n; i++) w[i] + = (Dep[i] << 1); Add.clear (); Del.clear ();    memset (tot, 0, sizeof (TOT));    for (int i = 1; I <= Q; i++) process (i, PS[I].C, Ps[i].s, Dep[ps[i].s]);            for (int i = 1, i <= N; i++) {for (int e = add.fir[i]; e; e = add.aft[e]) {int v = add.val[e];        tot[v]++;        } int u = pid[i];        Ans[u] + = Tot[w[u]];            for (int e = del.fir[i]; e; e = del.aft[e]) {int v = del.val[e];        tot[v]--;        }} for (int i = 1; I <= Q; i++) if (w[ps[i].c] = = Dep[ps[i].s]) ans[ps[i].c]--; for (int i = 1; I <= n; i++) printf ("%d%c", Ans[i], I < n?        ': ' \ n '); return 0;}

Day 1 T3 Classroom

I think of the problem in the examination room, but because the adjacency matrix in the edge of forgetting to remember Min is SB ....

Set F[0][I][J] indicates that the previous I request used J, the last one not taken the minimum desired distance; F[1][i][j] indicates that the first I request uses J Bar, the last one takes the minimum desired distance. Then, because each side of the line is independent, the transfer is directly multiplied by the probability of accumulation on the good.

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype    > #include <algorithm>using namespace Std;int read () {int x = 0, f = 1; char c = GetChar ();    while (!isdigit (c)) {if (c = = '-') f =-1; c = GetChar ();}    while (IsDigit (c)) {x = x * + C-' 0 '; c = GetChar ();} return x * f;} #define MAXN 2010#define maxv 310#define oo 1000000000int N, M, V, E, c[maxn], D[MAXN], D[MAXN][MAXN];d ouble P[MAXN], f[2] [MAXN]    [Maxn];void Up (double& A, double b) {a = min (a, b); return;}    int main () {freopen ("classroom.in", "R", stdin), Freopen ("Classroom.out", "w", stdout); n = read (); m = read (); v = read ();    E = Read ();    for (int i = 1; I <= n; i++) C[i] = read ();    for (int i = 1; I <= n; i++) D[i] = read ();    for (int i = 1; I <= n; i++) scanf ("%lf", &p[i]);        for (int i = 1; I <= v; i++) {d[i][i] = 0;    for (int j = i + 1; j <= V; j + +) D[i][j] = d[j][i] = oo; } for (int i = 1; I <= e;        i++) {int a = read (), B = Read (), C = Read ();        D[a][b] = min (d[a][b], c);    D[b][a] = min (D[b][a], c);                } for (int k = 1, k <= v; k++) for (int i = 1; I <= v; i++) for (int j = 1; J <= V; j + +)        D[i][j] = min (D[i][j], d[i][k] + d[k][j]);    for (int i = 0, I <= N; i++) for (int j = 0; J <= M; j + +) F[0][i][j] = f[1][i][j] = 1e9;    F[0][0][0] = 0.0; for (int i = 0; l <= N; i++) for (int j = 0; J <= min (i + 1, m); j + +) {double P = p[i], NP = p[i+1]            , p1 = 1.0-p, np1 = 1.0-NP;            Up (F[0][i+1][j], f[0][i][j] + d[c[i]][c[i+1]);            Up (F[1][i+1][j+1], F[0][i][j] + NP * D[c[i]][d[i+1]] + NP1 * d[c[i]][c[i+1]]);            Up (F[0][i+1][j], f[1][i][j] + P * d[d[i]][c[i+1]] + p1 * d[c[i]][c[i+1]]); Up (F[1][i+1][j+1], f[1][i][j] + p * NP * D[D[I]][D[I+1]] + p * NP1 * d[d[i]][c[i+1]] + p1 * NP * d[c[i]][d[i+1]] + p1 * NP       1 * d[c[i]][c[i+1]]); } double ans = 1e9;    for (int i = 0; I <= m; i++) up (ans, min (f[0][n][i], f[1][n][i]));        printf ("%.2lf\n", ans); return 0;}

Day 2 T1 problem

The recursive method is used to find the combined number and the real-time modulus K.

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype    > #include <algorithm>using namespace Std;int read () {int x = 0, f = 1; char c = GetChar ();    while (!isdigit (c)) {if (c = = '-') f =-1; c = GetChar ();}    while (IsDigit (c)) {x = x * + C-' 0 '; c = GetChar ();} return x * f;} #define MAXN 2010int C[2][MAXN], F[maxn][maxn];int main () {freopen ("problem.in", "R", stdin) freopen ("Problem.out", "W"    , stdout);    int size = 2000;    BOOL cur = 0;    int T = Read (), k = Read (); for (int i = 0; I <= size; i++, cur ^= 1) {c[cur][0] = 1;        C[cur][i] = 1;        for (int j = 1; j < I; j + +) C[cur][j] = (C[cur^1][j-1] + c[cur^1][j])% K; for (int j = 0; J <= I; j + +) {F[i][j] = (!            C[CUR][J]);            int t = 0;            if (j) f[i][j] + = f[i][j-1], t++;            if (J <= i-1) f[i][j] + = F[i-1][j], t++;      if (t = = 2 && i && j) f[i][j]-= f[i-1][j-1];  }} while (t--) {int n = read (), M = Read ();    printf ("%d\n", F[n][min (n,m)]); } return 0;}

Day 2 T2 Earthworm

Inspired by the data of q = 0, which does not increase the length of the earthworm, we find that the length of the two earthworms at each threshold must be monotonically reduced, so that O (n) can be done.

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype    > #include <algorithm>using namespace Std;int read () {int x = 0, f = 1; char c = GetChar ();    while (!isdigit (c)) {if (c = = '-') f =-1; c = GetChar ();}    while (IsDigit (c)) {x = x * + C-' 0 '; c = GetChar ();} return x * f;} #define MAXN 100010#define MAXM 7100010#define LL long longint N, M, q, u, V, T, A[maxn], B[MAXM], C[maxm], LB, RB, LC, RC , ANS[MAXM], Cnt;bool cmp (int a, int b) {return a > B;}    void process (int tmp, int i, int len) {int nb = (LL) tmp * u/v, NC = TMP-NB;    B[++RB] = nb-len-q;    C[++RC] = nc-len-q;    if (i% t = = 0) ans[++cnt] = tmp; return;}    int main () {freopen ("earthworm.in", "R", stdin), Freopen ("Earthworm.out", "w", stdout); n = read (); m = read (); Q = Read (); U = Read (); v = read ();    t = read ();    for (int i = 1; I <= n; i++) A[i] = read (); Sort (A + 1, a + n + 1, CMP);//for (int i = 1; I <= N;i++) printf ("%d%c", A[i], I < n?        ': ' \ n '); LB = 1; RB = 0; LC = 1;    rc = 0;    int PA = 1, len = 0;        for (int i = 1; I <= m; i++, len + = q) {int A, b, C; A = (PA <= n)?        A[PA] + len:-1; b = (lb <= rb)?        B[LB] + len:-1; c = (LC <= RC)?        C[LC] + len:-1;            if (a >= b && a >= c) {pa++;        Process (A, I, Len);            } else if (b >= a && b >= c) {lb++;        Process (b, I, Len);            } else {lc++;        Process (c, I, Len); }} for (int i = 1; I <= cnt; i++) printf ("%d%c", Ans[i], I < CNT?    ': ' \ n ');    if (!cnt) Putchar (' \ n ');    CNT = 0;        for (int i = 1; I <= n + m; i++) {int A, b, C; A = (PA <= n)?        A[PA] + len:-1; b = (lb <= rb)?        B[LB] + len:-1; c = (LC <= RC)?        C[LC] + len:-1;            if (a >= b && a >= c) {pa++; if (i% t = = 0) ANS[++CNT] = A;            } else if (b >= a && b >= c) {lb++;        if (i% t = = 0) ans[++cnt] = b;            } else {lc++;        if (i% t = = 0) ans[++cnt] = C; }} for (int i = 1; I <= cnt; i++) printf ("%d%c", Ans[i], I < CNT?    ': ' \ n ');        if (!cnt) Putchar (' \ n '); return 0;}

Day 2 T3 Angrybirds

Pressure DP, set F[s] means the minimum number of parabolic lines needed to kill the set S pig, the first pig to be found without being killed, and the other pig, two points to determine a parabola (0, 0), and then transfer to the current set and parabola through the collection of pigs set. Note that each parabolic pass through the pig's collection can be pretreated.

#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype > #include <algorithm> #include <cmath>using namespace Std;int read () {int x = 0, f = 1; char c = GetChar    ();    while (!isdigit (c)) {if (c = = '-') f =-1; c = GetChar ();}    while (IsDigit (c)) {x = x * + C-' 0 '; c = GetChar ();} return x * f;}    #define MAXN 23#define maxs 362154const Double eps = 1e-6;struct point {double x, y; Point () {}-point (Double _, double __): X (_), Y (__) {} BOOL operator < (const point& t) const {return x! = T. X? X < t.x:y < T.y; }} ps[maxn];int F[maxs], Ls[maxn][maxn];bool on (double A, double b, point P) {return fabs (A * p.x * p.x + b * p.x-p. y) <= EPS;}    void up (int& a, int b) {if (a < 0) A = b;    else a = min (a, b); return;}    int main () {freopen ("angrybirds.in", "R", stdin), Freopen ("Angrybirds.out", "w", stdout);    int T = read (); while (t--) {int n = read (); Read ();        for (int i = 0; i < n; i++) scanf ("%lf%lf", &ps[i].x, &AMP;PS[I].Y);        Sort (PS, PS + N);        Memset (F,-1, sizeof (f));        memset (LS, 0, sizeof (LS));        F[0] = 0; for (int i = 0; i < n; i++) for (int j = 0; J < N; j + +) if (i! = j) {Double x1 = ps[i].x, y1 = P                S[i].y, x2 = ps[j].x, y2 = ps[j].y;                Double b = (x1 * X1 * y2-x2 * x2 * y1)/(x1 * X1 * x2-x2 * x2 * x1);                Double A = (Y1-b * x1)/(x1 * x1);                if (a >= 0.0) continue;                int S = 0; for (int k = 0; k < n; k++) if (((S >> K & 1) ^ 1) && on (A, B, Ps[k])) S |= (1 <& Lt        k);        LS[I][J] = S;        } int all = (1 << N)-1; for (int S = 0; S <= All; s++) if (F[s] >= 0) for (int j = 0; J < N; j + +) if ((S >> J & 1) ^ 1) {int TS = S |                (1 << j);            Up (F[ts], F[s] + 1);    for (int i = j + 1; i < n; i++) if ((TS >> I & 1) ^ 1) up (F[ts | ls[i][j], f[s] + 1);            Break    } printf ("%d\n", F[all]); } return 0;}

Why this NOIP is the second problem most difficult TAT

NOIP2016 Topic Integration