HDU 2474

    這題關鍵是數據量太大，如果用普通的銀行家算法，最壞情況下時間複雜度爲O(N*N*M)，顯然會超時。正確的解法是對每種資源分別考慮，建立M個隊列。進程對每種資源的需求數量從小到大存入隊列，這樣每次只需檢查隊列頭部的進程即可。排序時間複雜度O(M*N*logN),每個進程至多出隊M次，時間複雜度爲O(N*M*M)。

    由於杭電的數據太弱，所以這個算法體現不出效果...

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <memory.h>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <iostream>
#include <sstream>

using namespace std;

int allo[3][50005], reqs[3][50005], avai[3], indx[3];
pair<int, int> queu[3][50005];
bool flag[3][50005];
int n, m;

int main() {
    int i, j, k, cunt;
    bool sign;
    while (scanf("%d %d", &n, &m) != EOF) {
        for (i = 0; i < m; i++)
            for (j = 0; j < n; j++)
                scanf("%d", &allo[i][j]);
        for (i = 0; i < m; i++)
            for (j = 0; j < n; j++) {
                scanf("%d", &reqs[i][j]);
                queu[i][j].first = reqs[i][j];
                queu[i][j].second = j;
            }
        for (i = 0; i < m; i++)
            scanf("%d", &avai[i]);
        for (i = 0; i < m; i++)
            sort(queu[i], queu[i] + n);
        sign = true;
        cunt = 0;
        memset(indx, 0, sizeof(indx));
        memset(flag, 0, sizeof(flag));
        while (sign) {
            sign = false;
            for (i = 0; i < m; i++) {
                for (j = indx[i]; j < n; j++) {
                    if (queu[i][j].first > avai[i])
                        break;
                    flag[i][queu[i][j].second] = true;
                    for (k = 0; k < m; k++) {
                        if (!flag[k][queu[i][j].second])
                            break;
                    }
                    if (k == m) {
                        sign = true;
                        cunt++;
                        for (k = 0; k < m; k++) {
                            avai[k] += allo[k][queu[i][j].second];
                            if (avai[k] > 1000000000)
                                avai[k] = 1000000000;
                        }
                    }
                }
                indx[i] = j;
            }
        }
        if (cunt == n) puts("Yes");
        else puts("No");
    }
    return 0;
}