[BZOJ1629][Usaco2007 Demo]Cow Acrobats

1629: [Usaco2007 Demo]Cow Acrobats

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 1027  Solved: 534 [Submit][Status][Discuss]

Description

Farmer John's N (1 <= N <= 50,000) cows (numbered 1..N) are planning to run away and join the circus.  Their hoofed feet prevent them from tightrope walking and swinging from the trapeze (and their last attempt at firing a cow out of a cannon met with a dismal failure). Thus, they have decided to practice performing acrobatic stunts. The cows aren't terribly creative and have only come up with one acrobatic stunt: standing on top of each other to form a vertical stack of some height.  The cows are trying to figure out the order in which they should arrange themselves within this stack. Each of the N cows has an associated weight (1 <= W_i <= 10,000) and strength (1 <= S_i <= 1,000,000,000).  The risk of a cow collapsing is equal to the combined weight of all cows on top of her (not including her own weight, of course) minus her strength (so that a stronger cow has a lower risk).  Your task is to determine an ordering of the cows that minimizes the greatest risk of collapse for any of the cows. //有三个头牛,下面三行二个数分别代表其体重及力量 //它们玩叠罗汉的游戏,每个牛的危险值等于它上面的牛的体重总和减去它的力量值,因为它要扛起上面所有的牛嘛. //求所有方案中危险值最大的最小

Input

* Line 1: A single line with the integer N. * Lines 2..N+1: Line i+1 describes cow i with two space-separated        integers, W_i and S_i.

Output

* Line 1: A single integer, giving the largest risk of all the cows in        any optimal ordering that minimizes the risk.

Sample Input

3
10 3
2 5
3 3

Sample Output

2

OUTPUT DETAILS:

Put the cow with weight 10 on the bottom. She will carry the other
two cows, so the risk of her collapsing is 2+3-3=2. The other cows
have lower risk of collapsing.
 
贪心
对于两只牛$i$和$j$,如果$w_i+s_i<w_j+s_j$那么一定$i$在$j$上面
证明可以先考虑两只牛相邻,再归纳
竟然跑到了BZOJ rank1 虽然是水题。。
 
1629 Accepted 6096 kb 20 ms C++/Edit 912 B
 
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; 
char buf[1000000], *ptr = buf - 1;
inline int readint(){
    int n = 0;
    char ch = *++ptr;
    while(ch < '0' || ch > '9') ch = *++ptr;
    while(ch <= '9' && ch >= '0'){
        n = (n << 1) + (n << 3) + ch - '0';
        ch = *++ptr; 
    }
    return n;
}
const int maxn = 50000 + 10;
struct Node{
    int w, s;
    Node(){}
    bool operator < (const Node &x) const {
        return w + s < x.w + x.s;
    }
}a[maxn];
int main(){
    fread(buf, sizeof(char), sizeof(buf), stdin);
    int N = readint();
    for(int i = 1; i <= N; i++){
        a[i].w = readint();
        a[i].s = readint();
    }
    sort(a + 1, a + N + 1);
    int sum = 0, ans = -66662333;
    for(int i = 1; i <= N; i++){
        ans = max(ans, sum - a[i].s);
        sum += a[i].w;
    }
    printf("%d\n", ans);
    return 0;
}

 

posted @ 2017-09-11 20:16  Elder_Giang  阅读(128)  评论(0编辑  收藏  举报