HDU 5045 Contest（状压DP或费用流）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5045

题意：N个人去做M道题，给出这N个人解这M道题的概率，且任意两人做题数不能相差1，问做出所有题的最大期望

思路：因为做题数不超过1的限制，这相当于是以N个人为一轮做N道题，做完M道为止，故而可以用dp[i][j]代表解决前i道题时状态为j，当j == (1 << n) - 1时清零。

同样因为N个人解N道问题，这就变成了二分图最大权匹配问题，跑(M / N) + 1次费用流也行

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <climits>
#include <functional>
#include <deque>
#include <string>
#include <bitset>
#include <ctime>

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;

const int maxn = 10010;
const int inf = 0x3f3f3f3f;

double p[20][1010], dp[1010][(1 << 10) + 20];

int main()
{
int t;
cin >> t;
for (int ca = 1; ca <= t; ca++)
{
printf("Case #%d: ", ca);

int n, m;
scanf("%d%d", &n, &m);

for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
scanf("%lf", &p[i][j]);

for (int i = 0; i <= m; i++)
for (int j = 0; j <= (1 << n); j++)
dp[i][j] = -1;

dp[0][0] = 0;
for (int j = 1; j <= m; j++)
for (int i = 0; i < (1 << n); i++)
{
if (dp[j - 1][i] == -1) continue;

for (int k = 1; k <= n; k++)
{
int s =  (1 << (k - 1));
if (i & s) continue;

s |= i;
if (s == (1 << n) - 1) s = 0;
dp[j][s] = max(dp[j][s], dp[j - 1][i] + p[k][j]);
}
}

double ans = 0;
for (int i = 0; i < (1 << n); i++)
ans = max(ans, dp[m][i]);
printf("%.5f\n", ans);
}
return 0;
}

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <climits>
#include <functional>
#include <deque>
#include <ctime>

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;

const int MAXN = 100;
const int MAXM = 50000;
const int INF = 0x3f3f3f3f;

typedef long long ll;

struct Edge
{
int to, next, cap, flow;
double cost;
} edge[MAXM];

int head[MAXN], tol;
int pre[MAXN];
double dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1

void init(int n)
{
N = n;
tol = 0;
memset(head, -1, sizeof(head));
}

void addedge(int u, int v, int cap, double cost)
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}

bool spfa(int s, int t)
{
queue<int>q;
for (int i = 0; i < N; i++)
{
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (edge[i].cap > edge[i].flow &&
dis[v] > dis[u] + edge[i].cost )
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if (!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if (pre[t] == -1) return false;
else return true;
}

//返回的是最大流，cost存的是最小费用
double minCostMaxflow(int s, int t, double &cost)
{
int flow = 0;
cost = 0;
while (spfa(s, t))
{
int Min = INF;
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
{
if (Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
{
edge[i].flow += Min;
edge[i ^ 1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
return flow;
}

double p[20][1010];

int main()
{
int t;
cin >> t;
for (int ca = 1; ca <= t; ca++)
{
printf("Case #%d: ", ca);

int n, m;
scanf("%d%d", &n, &m);

for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++)
scanf("%lf", &p[i][j]);

double ans = 0, res;
int s = 0, t = n * 2 + 1;
for (int k = 0; k < m / n; k++)
{
init(t + 1);
for (int i = 1; i <= n; i++)
{
addedge(s, i, 1, 0);
addedge(i + n, t, 1, 0);

for (int j = n + 1; j <= n * 2; j++)
{
addedge(i, j, 1, -p[i][(j - n) + k * n]);
}
}

minCostMaxflow(s, t, res);
ans += res;
}

init(t + 1);
for (int i = n + 1; i <= n + (m % n); i++)
addedge(i, t, 1, 0);
for (int i = 1; i <= n; i++)
{
addedge(s, i, 1, 0);

for (int j = n + 1; j <= n + (m % n); j++)
{
addedge(i, j, 1, -p[i][(j - n) + n * (m / n)]);
}
}

minCostMaxflow(s, t, res);
ans += res;
printf("%.5f\n", -ans);
}
return 0;
}