SGU 257 Debt 贪心 + 模拟

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题意：有一个人欠三个人分别p o s（1<=p o s<=10^5）块钱，现在这个人手里有n（1<=n<=10^5）块crystal，但是每块crystal在不同人看来是不一样的

价值（1块或者2块），现在问是否存在一种分配方案使得能还清3个人的钱。

题解：首先给不同的crystal排优先级，然后枚举满足三个人的优先级后贪心。

Sure原创，转载请注明出处

#include <iostream>#include <cstdio>#include <memory.h>#include <algorithm>#define MAX(a , b) ((a) > (b) ? (a) : (b))using namespace std;const int pp[9] = {8,1,2,4,3,5,6,7};const int maxn = 100002;struct node{    int val[3];    int id,pro;    bool operator < (const node &other) const    {        return pro < other.pro;    }}crystal[maxn];int belong[maxn];int a[3],rank[3],n,p,o,s;void read(){    char str[5];    scanf("%d",&n);    for(int i=0;i<n;i++)    {        scanf("%s",str);        int c = 0;        crystal[i].id = i;        for(int j=0;j<3;j++)        {            c <<= 1;            if(str[j] == 'B')            {                crystal[i].val[j] = 2;                c |= 1;            }            else crystal[i].val[j] = 1;        }        crystal[i].pro = pp[c];    }    sort(crystal , crystal + n);    return;}void out(){    for(int i=0;i<n;i++)    {        if(belong[i] == 0) printf("P");        else if(belong[i] == 1) printf("O");        else printf("S");    }    puts("");    return;}bool check(int xx,int yy,int zz){    a[0] = p;    a[1] = o;    a[2] = s;    rank[0] = xx;    rank[1] = yy;    rank[2] = zz;    for(int i=0;i<n;i++)    {        int j;        for(j=0;j<3;j++)        {            if(crystal[i].val[rank[j]] == 2 && a[rank[j]] > 1)            {                a[rank[j]] -= 2;                belong[crystal[i].id] = rank[j];                break;            }        }        if(j == 3)        {            for(j=0;j<3;j++)            {                if(a[rank[j]] > 0)                {                    a[rank[j]]--;                    belong[crystal[i].id] = rank[j];                    break;                }            }            if(j == 3) belong[crystal[i].id] = 0;        }    }    return (a[0] <= 0 && a[1] <= 0 && a[2] <= 0);}void solve(){    if(check(0,1,2)) {out();return;}    if(check(0,2,1)) {out();return;}    if(check(1,0,2)) {out();return;}    if(check(1,2,0)) {out();return;}    if(check(2,0,1)) {out();return;}    if(check(2,1,0)) {out();return;}    puts("no solution");    return;}int main(){    while(~scanf("%d %d %d",&p,&o,&s))    {        read();        solve();    }    return 0;}