题面

Sol

首先显然吃饭久的要排在前面

之后再来分配队伍，设\(f[i][j]\)表示到第\(i\)个人，\(A\)队伍要等\(j\)的最小吃完饭时间

那么就是一个简单的背包吧。。。

# include <bits/stdc++.h>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

# define Sqr(x) ((x) * (x))

using namespace std;

typedef long long ll;

const int _(205);

IL ll Read(){

    RG ll x = 0, z = 1; RG char c = getchar();

    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;

    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);

    return x * z;

}

int n, a[_], b[_], sum[_], f[_][_ * _], ans = 2e9, id[_];

IL bool Cmp(RG int x, RG int y){  return b[x] > b[y];  }

int main(RG int argc, RG char* argv[]){

	n = Read();

	for(RG int i = 1; i <= n; ++i) a[i] = Read(), b[i] = Read(), id[i] = i;

	sort(id + 1, id + n + 1, Cmp);

	for(RG int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + a[id[i]];

	Fill(f, 63); f[0][0] = 0;

	for(RG int i = 1; i <= n; ++i)

		for(RG int j = 0; j <= sum[i]; ++j){

			f[i][j] = max(f[i - 1][j], (sum[i] - j) + b[id[i]]);

			if(j >= a[id[i]]) f[i][j] = min(f[i][j], max(f[i - 1][j - a[id[i]]], j + b[id[i]]));

		}

	for(RG int i = 0; i <= sum[n]; ++i) ans = min(ans, f[n][i]);

	printf("%d\n", ans);

    return 0;

}

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Bzoj1899: [Zjoi2004]Lunch 午餐

题面

Sol