HDU - 6396 Swordsman(读写挂)

http://acm.hdu.edu.cn/showproblem.php?pid=6396

题意：

一个人有k个属性，有n个怪物每个也有k个属性和k个强化，如果各项属性都大于怪物的话，那么就可以杀死怪物，并且强化自己各自的属性

思路：

其实这道题不是太难，只是普通的读入挂不能用就很骚，必须得用fread加强的读入挂。

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <string.h>
#include <queue>
#include <stack>
#include <deque>
#include <stdlib.h>
#include <bitset>
#include <functional>

using namespace std;

#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define maxn 100005
#define eps 0.00000001
#define PI acos(-1.0)
#define M 1e9 + 7
//---____________---------__---__--_--_-_-_--_-_-_-__-_-_-___--_-_-_-
namespace fastIO {
#define BUF_SIZE 100000
    //fread -> read
    bool IOerror = 0;
    inline char nc() {
        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
        if(p1 == pend) {
            p1 = buf;
            pend = buf + fread(buf, 1, BUF_SIZE, stdin);
            if(pend == p1) {
                IOerror = 1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
    }
    inline void read(int &x) {
        char ch;
        while(blank(ch = nc()));
        if(IOerror) return;
        for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
    }
#undef BUF_SIZE
};
using namespace fastIO;


//--________________-___________________________________//
struct XK {
    int val, id;
    XK(){}
    XK(int _val, int _id){
        val = _val; id = _id;
    }
    bool operator > (const XK &a) const {
        return val > a.val;
    }
    bool operator < (const XK &a) const {
        return val < a.val;
    }
};


struct SX {
    int w[10], v[10];
};

priority_queue<XK, vector<XK>, greater<XK> > que[6];
int NanZ[10], hurt[maxn];
SX G[maxn];


int main(int argc, const char * argv[]) {
    //freopen("/Volumes/mac 盘/Code/in.txt", "r", stdin);
    int T;
    read(T);
    while(T--) {
        int n, k;
        read(n); read(k);
        memset(hurt, 0, sizeof(hurt));
        memset(G, 0, sizeof(G));
        for (int i = 0; i < k; i ++)
            read(NanZ[i]);
        for (int i = 0; i < n; i ++) {
            for (int j = 0; j < k; j ++) {
                read(G[i].w[j]);
                que[j].push(XK(G[i].w[j], i));
            }
            for (int j = 0; j < k; j ++)
                read(G[i].v[j]);
        }
        int ans = 0;
        bool Flag = 1;
        while(Flag) {
            Flag = 0;
            for (int i = 0; i < k; i ++) {
                while(!que[i].empty()) {
                    XK t = que[i].top();
                    if(t.val <= NanZ[i]) {
                        hurt[t.id] ++;
                        que[i].pop();
                        if(hurt[t.id] == k) {
                            for (int i = 0; i < k; i ++) 
                                NanZ[i] += G[t.id].v[i];
                            ans ++;
                            Flag = 1;
                        }
                    }else
                        break;
                }
            }
        }
        printf("%d\n", ans);
        for (int i = 0; i < k; i ++)
            printf("%d%c", NanZ[i], i == k - 1? '\n' : ' ');

        for (int i = 0; i < 6; i ++)
            while(!que[i].empty())
                que[i].pop();
    }
    return 0;
}