BZOJ3523 [Poi2014]Bricks 【贪心】

题目链接

BZOJ3523

题解

简单的贪心题
优先与上一个不一样且数量最多的,如果有多个相同,则优先选择非结尾颜色
比较显然,但不知怎么证

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) (cp){a,b}
#define cls(s) memset(s,0,sizeof(s))
#define LL long long int
using namespace std;
const int maxn = 1000005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int ans[maxn],n,s,t,m;
struct cp{
    int cnt,id;
};
inline bool operator <(const cp& a,const cp& b){
    return a.cnt == b.cnt ? a.id != t : a.cnt < b.cnt;
}
priority_queue<cp> q;
int main(){
    n = read(),s = read(),t = read();
    if (n == 1 && (s != t)){puts("0"); return 0;}
    int tmp;
    REP(i,n){
        tmp = read(); m += tmp;
        if (i == s) tmp--;
        if (i == t) tmp--;
        if (tmp == -1){puts("0"); return 0;}
        if (tmp) q.push(mp(tmp,i));
    }
    cp u,v;
    ans[1] = s; ans[m] = t;
    for (int i = 2; i < m; i++){
        u = q.top(); q.pop();
        if (u.id == ans[i - 1]){
            if (q.empty()){puts("0"); return 0;}
            v = q.top(); q.pop();
            ans[i] = v.id; v.cnt--;
            q.push(u);
            if (v.cnt) q.push(v);
        }
        else {
            ans[i] = u.id; u.cnt--;
            if (u.cnt) q.push(u);
        }
    }
    if (ans[m - 1] == t){puts("0"); return 0;}
    for (int i = 1; i <= m; i++)
        printf("%d ",ans[i]);
    return 0;
}

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转载自www.cnblogs.com/Mychael/p/9094419.html