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; }
时间: 2024-10-27 01:50:43