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[Offer收割]编程练习赛32

时间:2017-10-27 16:03:08      阅读:179      评论:0      收藏:0      [点我收藏+]

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 气泡图

两两判断关系,dfs。

技术分享
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
double x[1005], y[1005], r[1005];
int f[1005][1005];
const double eps = 1e-10;
#include<vector>
using namespace std;
vector<int> G[1005];
int p[1005], father[1005];
void dfs(int x, int fa) {
    father[x] = fa;
    for (int i = 0; i < G[x].size(); i++) {
        p[G[x][i]]--;
    }
    for (int i = 0; i < G[x].size(); i++) {
        int v = G[x][i];
        if (p[v] == 0 && father[v]==0) {
            dfs(v, x);
        }
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%lf%lf%lf", &x[i], &y[i], &r[i]);
    }
    memset(f, 0, sizeof(f));
    memset(p, 0, sizeof(p));
    memset(father, 0, sizeof(father));
    for (int i = 1; i <= n; i++) {
        for (int j = i + 1; j <= n; j++) {
            if (i != j) {
                double d = sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
                if (r[i] - r[j] >= d) {
                    f[i][j] = 1;
                    G[i].push_back(j);
                    f[j][i] = -1;
                    p[j]++;
                }
                if (r[j] - r[i] >= d) {
                    f[i][j] = -1;
                    f[j][i] = 1;
                    G[j].push_back(i);
                    p[i]++;
                }
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        if (p[i] == 0) {
            dfs(i, 0);
            break;
        }
    }
    for (int i = 1; i <= n; i++) {
        printf("%d\n", father[i]);
    }
    return 0;
}
View Code

候选人追踪

维护S集合中的最小值和其余候选人中的最大值

技术分享
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
template <class T> class SegmentTree {
public:
    T dat, lazy;
    int leftBorder, rightBorder, mid;
    SegmentTree * leftSon, * rightSon;
    T(* lazyfunc)(T, T);
    T(* mergefunc)(T, T);
    SegmentTree() {
        leftBorder = rightBorder = -1;
        leftSon = rightSon = NULL;
    }
    void pushdown();
    void pushup();
    void Build(T *, int, int, T(*)(T, T), T(*)(T, T));
    void Modify(int, int, T);
    T Query(int, int);
    void Free();
};
template<class T> void SegmentTree<T>::pushdown() {
    if (lazy && leftBorder != rightBorder) {
        leftSon->dat = lazyfunc(leftSon->dat, lazy);
        rightSon->dat = lazyfunc(rightSon->dat, lazy);
        leftSon->lazy = lazyfunc(leftSon->lazy, lazy);
        rightSon->lazy = lazyfunc(rightSon->lazy, lazy);
    }
    lazy = (T)0;
}
template<class T> void SegmentTree<T>::pushup() {
    dat = mergefunc(leftSon->dat, rightSon->dat);
}
template<class T> void SegmentTree<T>::Build(T * S, int l, int r, T(* lfunc)(T, T), T(* mfunc)(T, T)) {
    if (l > r) {
        return;
    }
    lazy = (T)0;
    leftBorder = l;
    rightBorder = r;
    mid = (leftBorder + rightBorder) >> 1;
    lazyfunc = lfunc;
    mergefunc = mfunc;
    if (l == r) {
        dat = S[l];
        return;
    }
    leftSon = new SegmentTree;
    leftSon->Build(S, l, mid, lfunc, mfunc);
    rightSon = new SegmentTree;
    rightSon->Build(S, mid + 1, r, lfunc, mfunc);
    pushup();
}
template<class T> void SegmentTree<T>::Modify(int l, int r, T NewDat) {
    if (l > r || l < leftBorder || rightBorder < r) {
        return;
    }
    if (leftBorder == l && rightBorder == r) {
        dat = lazyfunc(dat, NewDat);
        lazy = lazyfunc(lazy, NewDat);
        return;
    }
    pushdown();
    if (r <= mid) {
        leftSon->Modify(l, r, NewDat);
    } else if (mid < l) {
        rightSon->Modify(l, r, NewDat);
    } else {
        leftSon->Modify(l, mid, NewDat);
        rightSon->Modify(mid + 1, r, NewDat);
    }
    pushup();
}
template<class T> T SegmentTree<T>::Query(int l, int r) {
    if (l > r || l < leftBorder || rightBorder < r) {
        return dat;
    }
    pushdown();
    if (l == leftBorder && r == rightBorder) {
        return dat;
    }
    if (r <= mid) {
        return leftSon->Query(l, r);
    } else if (mid < l) {
        return rightSon->Query(l, r);
    } else {
        return mergefunc(leftSon->Query(l, mid), rightSon->Query(mid + 1, r));
    }
}
template<class T> void SegmentTree<T>::Free() {
    if (leftSon != NULL) {
        leftSon->Free();
    }
    if (rightSon != NULL) {
        rightSon->Free();
    }
    delete leftSon;
    delete rightSon;
}
int lazyfunc(int a, int b) {
    return a + b;
}
int mergefunc(int a, int b) {
    return a > b ? b : a;
}
int n, k;
bool f[320000];
class tick {
public:
    int t, c;
};
int cmp(const void *x, const void *y) {
    tick * tx = (tick *)x;
    tick * ty = (tick *)y;
    return tx->t > ty->t ? 1 : -1;
}
tick p[320000];
int index_____[320000], a[320000];
SegmentTree<int> st;
int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    memset(a, 0, sizeof(a));
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++) {
        scanf("%d%d", &p[i].t, &p[i].c);
    }
    memset(f, false, sizeof(f));
    for (int i = 0; i < k; i++) {
        int s;
        scanf("%d", &s);
        f[s] = true;
    }
    int cnt = 0;
    for (int i = 0; i < 315159; i++) {
        if (f[i]) {
            index_____[i] = cnt++;
        }
    }
    qsort(p, n, sizeof(tick), cmp);
    int ptr = 0, ans = 0, max = 0, time = p[0].t, s = -1;
    st.Build(a, 0, k - 1, lazyfunc, mergefunc);
    while (ptr < n) {
        do {
            if (f[p[ptr].c]) {
                st.Modify(index_____[p[ptr].c], index_____[p[ptr].c], 1);
            } else {
                a[p[ptr].c]++;
                if (a[p[ptr].c] > max) {
                    max = a[p[ptr].c];
                }
            }
            ptr++;
        } while (p[ptr].t == p[ptr - 1].t && ptr < n);
        int q = st.Query(0, k - 1);
        if (s == 1) {
            ans += p[ptr - 1].t - time;
        }
        time = p[ptr - 1].t;
        if (q > max) {
            s = 1;
        } else {
            s = -1;
        }
    }
    printf("%d\n", ans);
    return 0;
}
View Code

 

[Offer收割]编程练习赛32

标签:click   main   gif   namespace   else   family   mod   bsp   int   

原文地址:http://www.cnblogs.com/dramstadt/p/7742752.html

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