【HDOJ】2333 Assemble

二分+贪心策略。其中注释处很重要。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;

#define MAXL 25
#define MAXN 1005
#define INF  1000000000

typedef struct {
    char name[MAXL];
    int p, q;
} comp_st;

comp_st comps[MAXN];
int t, n, b;
int ns[MAXN], nn;

bool comp(comp_st a, comp_st b) {
    int tmp = strcmp(a.name, b.name);
    if (tmp == 0)
        return a.q > b.q;
    else
        return tmp < 0;
}

bool chk(int x) {
    int i, j;
    int sum = 0, min;

    for (i=0; i<nn; ++i) {
        min = INF;
        for (j=ns[i]; j<ns[i+1]; ++j) {
            if (comps[j].q>=x && min>comps[j].p)
                min = comps[j].p;
        }
        /* pay attention to here.*/
        if (min == INF)
            return false;
        sum += min;
    }
    return sum<=b;
}

int main() {
    int i, min, max, mid, v;
    char str[MAXL];

    scanf("%d", &t);

    while (t--) {
        scanf("%d %d", &n, &b);
        min = INF;
        max = 0;
        for (i=0; i<n; ++i) {
            scanf("%s %s %d %d", comps[i].name, str, &comps[i].p, &comps[i].q);
            if (comps[i].q < min) min = comps[i].q;
            if (comps[i].q > max) max = comps[i].q;
        }
        sort(comps, comps+n, comp);
        ns[0] = 0;
        for (nn=0,i=1; i<n; ++i) {
            if (strcmp(comps[i-1].name, comps[i].name) != 0)
                ns[++nn] = i;
        }
        ns[++nn] = n;
        v = 0;
        while (min <= max) {
            mid = (min+max)/2;
            if ( chk(mid) ) {
                v = mid;
                min = mid+1;
            } else {
                max = mid-1;
            }
        }
        printf("%d
", v);
    }

    return 0;
}