LightOJ 1350 Aladdin and the Grand Feast 最大流（难）

题目：http://www.lightoj.com/volume_showproblem.php?problem=1350

题意：要举办宴会，时长为e，有t个桌子，每个桌子可以坐c个人。每个宾客吃饭的时间为[a, b)，要吃f单位的食物，没人每个单位时间只能吃一单位食物。问能不能满足所有宾客的需求。若能，则输出每个单位时间桌子上的宾客，多解任意输出一个

思路：常规思路是把宾客和每个时间单位作为点，很好套路，但单位时间为1e4，肯定会T。我们只用每个宾客吃饭的时间[a, b)这两个时间点，这样点就不超过200个，但是后续处理挺恶心的。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define debug() puts("here");
using namespace std;

const int N = 210, M = 10010;
const int INF = 0x3f3f3f3f;
struct edge
{
int to, cap, next;
} g[N*N*2];
int cnt, nv, head
, level
, gap
, cur
, pre
;
int aa
, bb
, cc
, deg[M], id[M], last[M], arr[M];
char ans[M][60];
int cas;
void add_edge(int v, int u, int cap)
{
g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;
g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++;
}
int sap(int s, int t)
{
memset(level, 0, sizeof level);
memset(gap, 0, sizeof gap);
memcpy(cur, head, sizeof head);
gap[0] = nv;
int v = pre[s] = s, flow = 0, aug = INF;
while(level[s] < nv)
{
bool flag = false;
for(int &i = cur[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && level[v] == level[u] + 1)
{
flag = true;
pre[u] = v;
v = u;
aug = min(aug, g[i].cap);
if(v == t)
{
flow += aug;
while(v != s)
{
v = pre[v];
g[cur[v]].cap -= aug;
g[cur[v]^1].cap += aug;
}
aug = INF;
}
break;
}
}
if(flag) continue;
int minlevel = nv;
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && minlevel > level[u])
minlevel = level[u], cur[v] = i;
}
if(--gap[level[v]] == 0) break;
level[v] = minlevel + 1;
gap[level[v]]++;
v = pre[v];
}
return flow;
}
bool cmp(int a, int b)
{
return deg[a] > deg[b];
}
char work(int i)
{
if(i <= 26) return i - 1 + 'a';
else return i - 26 - 1 + 'A';
}
int main()
{
int T, n, t, c, e;
scanf("%d", &T);
while(T--)
{
cnt = 0;
memset(head, -1, sizeof head);
scanf("%d%d%d%d", &n, &t, &c, &e);
int tot = 0, sum = 0;
for(int i = 1; i <= n; i++)
{
scanf("%d%d%d", &aa[i], &bb[i], &cc[i]);
arr[tot++] = aa[i], arr[tot++] = bb[i];
sum += cc[i];
}
if(sum > e * t * c)
{
printf("Case %d: No\n", ++cas); continue;
}
sort(arr, arr + tot);
tot = unique(arr, arr + tot) - arr;
arr[tot] = arr[tot-1] + 1;
for(int i = 1; i <= n; i++)
{
aa[i] = lower_bound(arr, arr + tot, aa[i]) - arr + 1;
bb[i] = lower_bound(arr, arr + tot, bb[i]) - arr + 1;
}
int ss = 0, tt = n + tot + 1;
for(int i = 1; i <= n; i++) add_edge(ss, i, cc[i]);
for(int i = 0; i < tot; i++) add_edge(n + i + 1, tt, (arr[i+1] - arr[i]) * t * c);
for(int i = 1; i <= n; i++)
for(int j = aa[i]; j < bb[i]; j++)
add_edge(i, n + j, arr[j] - arr[j-1]);
nv = tt + 1;
int res = sap(ss, tt);
if(res != sum)
{
printf("Case %d: No\n", ++cas); continue;
}
printf("Case %d: Yes\n", ++cas);
memset(last, 0, sizeof last);
for(int i = 1; i <= e; i++)
for(int j = 0; j <= t * c; j++) ans[i][j] = '.';
for(int i = 1; i <= n; i++)
for(int j = head[i]; j != -1; j = g[j].next)
if(g[j^1].cap)
{
int l = g[j].to - n;
for(int k = arr[l-1]; k < arr[l]; k++)
deg[k] = t * c - last[k], id[k] = k;
sort(id + arr[l-1], id + arr[l], cmp);
for(int k = arr[l-1]; k < g[j^1].cap + arr[l-1]; k++)
{
int x = id[k];
ans[x][last[x]++] = work(i);
}
}
for(int i = 1; i < e; i++)
for(int j = 0; j < t * c; j++)
{
if(j && j % c == 0) printf("|");
printf("%c", ans[i][j]);
if(j + 1 == t * c) printf("\n");
}
}
return 0;
}