标签:read zjoi dig targe inline cst class memset http
显然吃饭时间越长的人排在前面越好,所以先排序。
f[i][j]表示前i个人,A队的打饭时间为j的最优解,每个人只有两种选择,去A队或是去B队。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 201
#define max(x, y) ((x) > (y) ? (x) : (y))
#define min(x, y) ((x) < (y) ? (x) : (y))
int n, ans = ~(1 << 31);
int f[N][40001], sum[N];
//f[i][j]表示前i个人,A队打饭时间为j的最优解
struct node
{
int a, b;
}p[N];
inline int read()
{
int x = 0, f = 1;
char ch = getchar();
for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
return x * f;
}
inline bool cmp(node x, node y)
{
return x.b > y.b;
}
int main()
{
int i, j, k;
n = read();
for(i = 1; i <= n; i++)
{
p[i].a = read();
p[i].b = read();
}
std::sort(p + 1, p + n + 1, cmp);
for(i = 1; i <= n; i++) sum[i] = sum[i - 1] + p[i].a;
memset(f, 127 / 3, sizeof(f));
f[0][0] = 0;
for(i = 1; i <= n; i++)
for(j = 0; j <= sum[i - 1]; j++)
{
f[i][j] = min(f[i][j], max(f[i - 1][j], sum[i - 1] - j + p[i].a + p[i].b));
f[i][j + p[i].a] = min(f[i][j + p[i].a], max(f[i - 1][j], j + p[i].a + p[i].b));
}
for(i = 0; i <= sum[n]; i++) ans = min(ans, f[n][i]);
printf("%d\n", ans);
return 0;
}
标签:read zjoi dig targe inline cst class memset http
原文地址:http://www.cnblogs.com/zhenghaotian/p/7347938.html