題目鏈接
想了許久,一看題解是一種比較sao的辦法.
1.源點連第一天
匯點連最後一天
容量爲INF費用爲0
2.然後每一天向後一天連一條容量爲INF-a[i]
費用爲0的邊
3.然後將每一類志願者s[i]與t[i]+1連一條容量爲
INF花費爲c[i]的邊
這樣爲了保證最大流是inf,圖會從3類邊中花費補流,最後最小費即爲所求.
下面是ac代碼:
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>
#include <cstdio>
#include <cmath>
#define ll long long
using namespace std;
#define ll long long
#define _min(x, y) ((x)>(y)?(y):(x))
#define pr(x, y) make_pair((x),(y))
const int M = 2e7;
const int N = 1e4;
const int inf = 0x3f3f3f3f;
int tot;
int mx, h[N], dis[N], pre[N];
int he[N], ne[M], ver[M], e[M], c[M];
void init(int n)
{
mx = n;
tot = 1;
}
void add(int x, int y, int w, int _c)
{
ver[++tot] = y;
ne[tot] = he[x];
e[tot] = w;
c[tot] = _c;
he[x] = tot;
ver[++tot] = x;
ne[tot] = he[y];
e[tot] = 0;
c[tot] = -_c;
he[y] = tot;
}
int EK(int s, int t, int f, int &flow)
{
int res = 0;
fill(h, h + 1 + mx, 0);
while (f)
{
priority_queue <pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;
fill(dis, dis + 1 + mx, inf);
dis[s] = 0; q.push(pair<int, int>(0, s));
while (q.size())
{
pair<int, int> now = q.top(); q.pop();
int te = now.second;
if (dis[te] < now.first)continue;
for (int i = he[te]; i; i = ne[i])
{
if (e[i] > 0 && dis[ver[i]] > dis[te] + c[i] + h[te] - h[ver[i]])
{
dis[ver[i]] = dis[te] + c[i] + h[te] - h[ver[i]];
pre[ver[i]] = i;
q.push(pair<int, int>(dis[ver[i]], ver[i]));
}
}
}
if (dis[t] == inf) break;
for (int i = 0; i <= mx; ++i) h[i] += dis[i];
int d = f;
for (int i = pre[t]; i; i = pre[ver[i^1]]) d = min(d, e[i]);
f -= d; flow += d; res += d*h[t];
for (int i = pre[t]; i; i = pre[ver[i^1]])
{
e[i] -= d;
e[i^1] += d;
}
}
return res;
}
int main()
{
int n, m;
cin >> n >> m;
int s = 0, t = n +2;
init(t);
for (int i = 1; i <= n; i++)
{
int te; scanf("%d", &te);
add(i, i + 1, inf - te, 0);
}
add(s, 1, inf, 0);
add(n+1, t, inf, 0);
for (int i = 1; i <= m; i++)
{
int l, r, c;
scanf("%d%d%d", &l, &r, &c);
add(l, r+1, inf, c);
}
int flow = 0;
int ans = EK(s, t, inf, flow);
printf("%d\n", ans);
return 0;
}