首页 > 代码库 > POJ3614 Sunscreen 【贪心】

POJ3614 Sunscreen 【贪心】

2024-08-02 10:18:45   219人阅读

Sunscreen
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 4204 Accepted: 1458

Description

To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide with sunscreen when they‘re at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPFi ≤ 1,000; minSPFi≤ maxSPFi ≤ 1,000) that will work. If the SPF rating is too low, the cow suffers sunburn; if the SPF rating is too high, the cow doesn‘t tan at all........

The cows have a picnic basket with L (1 ≤ L ≤ 2500) bottles of sunscreen lotion, each bottle i with an SPF rating SPFi (1 ≤ SPFi ≤ 1,000). Lotion bottle i can cover coveri cows with lotion. A cow may lotion from only one bottle.

What is the maximum number of cows that can protect themselves while tanning given the available lotions?

Input

* Line 1: Two space-separated integers: C and L
* Lines 2..C+1: Line i describes cow i‘s lotion requires with two integers: minSPFi and maxSPFi 
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPFi and coveri

Output

A single line with an integer that is the maximum number of cows that can be protected while tanning

Sample Input

3 2
3 10
2 5
1 5
6 2
4 1

Sample Output

2

Source

USACO 2007 November Gold

贪心解法：将牛和防晒霜分别按照first值升序排序，然后对于每个防晒霜，将起点<=它的位置的牛都放入优先队列里，然后从优先队列里取出右端值最小的合法牛加入ans。0ms。

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
#include <iostream>

#define maxn 2510
using namespace std;
struct Node {
    int first, second;
    friend bool operator<(const Node& a, const Node& b) {
        return a.second > b.second;
    }
} cow[maxn], sun[maxn];

bool cmp(const Node& a, const Node& b) {
    return a.first < b.first;
}

int main() {
    int C, L, i, j, ans = 0;
    Node tmp;
    scanf("%d%d", &C, &L);
    for(i = 0; i < C; ++i)
        scanf("%d%d", &cow[i].first, &cow[i].second);
    for(i = 0; i < L; ++i)
        scanf("%d%d", &sun[i].first, &sun[i].second);
    sort(cow, cow + C, cmp);
    sort(sun, sun + L, cmp);
    priority_queue<Node> PQ;
    for(i = j = 0; i < L; ++i) {
        while(j < C) {
            if(cow[j].first > sun[i].first) break;
            PQ.push(cow[j++]);
        }
        while(!PQ.empty() && sun[i].second) {
            tmp = PQ.top();
            if(tmp.second < sun[i].first)
                PQ.pop();
            else if(tmp.first <= sun[i].first) {
                --sun[i].second;
                ++ans; PQ.pop();
            }
        }
    }

    printf("%d\n", ans);
    return 0;
}

网络流解法：比赛时用这个代码险过，900多ms。


#include <stdio.h>
#include <string.h>

#define maxn 4010
#define maxm maxn * 2000
#define inf 0x3f3f3f3f

int head[maxn], id, N, L, source, sink;
struct Node {
    int u, v, c, next;
} E[maxm];

void addEdge(int u, int v, int c) {
    E[id].u = u; E[id].v = v;
    E[id].c = c; E[id].next = head[u];
    head[u] = id++;

    E[id].u = v; E[id].v = u;
    E[id].c = 0; E[id].next = head[v];
    head[v] = id++;
}

void getMap() {
    memset(head, -1, sizeof(head));
    int i, j, a, b; 
    source = 3502; sink = source + 1;
    for(i = 0; i < N; ++i) {
        addEdge(source, i, 1);
        scanf("%d%d", &a, &b);
        for(j = a; j <= b; ++j)
            addEdge(i, j + N, 1);
    }
    while(L--) {
        scanf("%d%d", &a, &b);
        addEdge(N + a, sink, b);
    }
}

int dep[maxn], ps[maxn], cur[maxn];
int Dinic(int n, int s, int t) {
    int tr, res = 0;
    int i, j, k, f, r, top;
    while(true) {
        memset(dep, -1, n * sizeof(int));
        for(f = dep[ps[0] = s] = 0, r = 1; f != r; )
            for(i = ps[f++], j = head[i]; j != -1; j = E[j].next) {
                if(E[j].c && -1 == dep[k=E[j].v]) {
                    dep[k] = dep[i] + 1; ps[r++] = k;
                    if(k == t) {
                        f = r; break;
                    }
                }
            }
        if(-1 == dep[t]) break;

        memcpy(cur, head, n * sizeof(int));
        for(i = s, top = 0; ; ) {
            if(i == t) {
                for(k = 0, tr = inf; k < top; ++k)
                    if(E[ps[k]].c < tr) tr = E[ps[f=k]].c;
                for(k = 0; k < top; ++k)
                    E[ps[k]].c -= tr, E[ps[k]^1].c += tr;
                res += tr; i = E[ps[top = f]].u;
            }
            for(j = cur[i]; cur[i] != -1;j = cur[i] = E[cur[i]].next)
                if(E[j].c && dep[i] + 1 == dep[E[j].v]) break;
            if(cur[i] != -1) {
                ps[top++] = cur[i];
                i = E[cur[i]].v;
            } else {
                if(0 == top) break;
                dep[i] = -1; i = E[ps[--top]].u;
            }
        }
    }
    return res;
}

int main() {
    // freopen("stdin.txt", "r", stdin);
    scanf("%d%d", &N, &L);
    getMap();
    printf("%d\n", Dinic(sink + 1, source, sink));
    return 0;
}


POJ3614 Sunscreen 【贪心】


联系
我们