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-1void 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;}