題目鏈接:BZOJ1070
題目大意
啊,好懶啊,中文題面以後就不說了吧。
分析
1. 感覺上這是一道網絡流的題,卻想不出來怎麼建圖;其實這道題的建圖思路很巧妙。
2. 對於一個工作人員
3. 把每個工作人員拆成
4.
5. 跑一遍費用流,答案除以
上代碼
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 10 + 5;
const int M = 60 + 5;
const int INF = 0x3f3f3f3f;
int n, m;
int S, T, cnt;
int C[N][M], id1[N][M], id2[M];
int head[N * M], len;
struct nodeLib {
int to, nxt, val, flow;
inline void add(int a, int b, int c, int d) {
to = b, val = d, flow = c;
nxt = head[a], head[a] = len++;
}
} lib[N * M * M << 1];
inline int read() {
char ch;
int ans = 0, neg = 1;
while (ch = getchar(), ch < '0' || ch > '9')
if (ch == '-') neg = -1;
while (ch >= '0' && ch <= '9')
ans = ans * 10 + ch - '0', ch = getchar();
return ans * neg;
}
inline void makePath(int a, int b, int c, int d) {
lib[len].add(a, b, c, d);
lib[len].add(b, a, 0, -d);
}
void init() {
n = read(), m = read();
len = 2, S = ++cnt, T = ++cnt;
for (int j = 1; j <= m; j++)
for (int i = 1; i <= n; i++)
C[i][j] = read();
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
makePath(S, id1[i][j] = ++cnt, 1, 0);
for (int i = 1; i <= m; i++)
makePath(id2[i] = ++cnt, T, 1, 0);
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
for (int k = 1; k <= m; k++)
makePath(id1[i][j], id2[k], 1, j * C[i][k]);
}
queue <int> Q;
bool inQ[N * M];
int dist[N * M], preE[N * M], preV[N * M];
bool SPFA() {
memset(dist, 0x3f, sizeof(dist));
dist[S] = 0, Q.push(S), inQ[S] = true;
while (!Q.empty()) {
int tmp = Q.front();
Q.pop(), inQ[tmp] = false;
for (int p = head[tmp]; p; p = lib[p].nxt) {
int now = lib[p].to, val = lib[p].val;
if (lib[p].flow && dist[now] > dist[tmp] + val) {
dist[now] = dist[tmp] + val;
preE[now] = p, preV[now] = tmp;
if (!inQ[now]) Q.push(now), inQ[now] = true;
}
}
}
return dist[T] != INF;
}
int MCMF() {
int ans = 0;
while (SPFA()) {
int maxf = INF;
for (int p = T; p != S; p = preV[p])
maxf = min(maxf, lib[preE[p]].flow);
for (int p = T; p != S; p = preV[p])
lib[preE[p]].flow -= maxf, lib[preE[p] ^ 1].flow += maxf;
ans += dist[T] * maxf;
}
return ans;
}
int main() {
init();
printf("%.2lf\n", (double)MCMF() / m);
return 0;
}
我還是naive啊,沒有想出來建圖。
以上