初见安~这里是传送门:洛谷P4041 分配问题
题解
【怎么这么水的题都要写题解啊。】
每个人选一个工作,求最大最小收益,跑网络最大/最小流就行了。费用放中间,流量限制放两边。
题目过水所以看代码建图就行了。费用流都是模板操作
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define maxn 500
#define maxm 200005
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
int read() {
int x = 0, f = 1, ch = getchar();
while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
while(isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
return x * f;
}
struct edge {int to, w, f, nxt;} e[maxm], e2[maxm];
int head[maxn], k = 0;
void add(int u, int v, int w, int f) {
e[k] = {v, w, f, head[u]}; head[u] = k++;
e[k] = {u, 0, -f, head[v]}; head[v] = k++;
}
int S, T;
int dis[maxn], pre[maxn], incf[maxn];
bool vis[maxn];
bool spfa() {
memset(dis, 0x3f, sizeof dis); dis[S] = 0;
memset(vis, 0, sizeof vis); vis[S] = true;
memset(incf, 0x3f, sizeof incf); incf[S] = INF;
queue<int> q; q.push(S);
while(q.size()) {
register int u = q.front(), v; q.pop(); vis[u] = false;
for(int i = head[u]; ~i; i = e[i].nxt) {
v = e[i].to; if(e[i].w && dis[u] + e[i].f < dis[v]) {
dis[v] = dis[u] + e[i].f;
incf[v] = min(incf[u], e[i].w); pre[v] = i;
if(!vis[v]) q.push(v), vis[v] = true;
}
}
}
if(dis[T] == INF) return false; return true;
}
ll ans = 0;
void update() {
register int u = T, i;
while(u != S) {
i = pre[u];
e[i].w -= incf[T], e[i ^ 1].w += incf[T];
u = e[i ^ 1].to;
}
ans += 1ll * incf[T] * dis[T];
}
int n;
signed main() {
memset(head, -1, sizeof head);
n = read(); S = 0, T = n + n + 1;
for(int i = 1; i <= n; i++) {//建图,n+1~2n是任务
add(S, i, 1, 0); add(i + n, T, 1, 0);
for(int j = 1, x; j <= n; j++) x = read(), add(i, j + n, INF, x);
}
for(int i = 0; i < k; i++) e2[i] = e[i];
while(spfa()) update();
printf("%lld\n", ans); ans = 0;
for(int i = 0; i < k; i++) e[i] = e2[i], e[i].f = -e[i].f;//建相反数图跑最小费用=最大费用相反数
while(spfa()) update();
printf("%lld\n", -ans);
return 0;
}
迎评:)
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