hdu 5045 Contest--2014acm上海赛区网络赛
来源：程序员人生 发布时间：2014-10-12 20:37:02 阅读次数：3537次
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5045
Contest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 171 Accepted Submission(s): 60
Problem Description
In the ACM International Collegiate Programming Contest, each team consist of three students. And the teams are given 5 hours to solve between 8 and 12 programming problems.
On Mars, there is programming contest, too. Each team consist of N students. The teams are given M hours to solve M programming problems. Each team can use only one computer, but they can’t cooperate to solve a problem. At the beginning of the ith hour, they
will get the ith programming problem. They must choose a student to solve this problem and others go out to have a rest. The chosen student will spend an hour time to program this problem. At the end of this hour, he must submit his program. This program is
then run on test data and can’t modify any more.
Now, you have to help a team to find a strategy to maximize the expected number of correctly solved problems.
For each problem, each student has a certain probability that correct solve. If the ith student solve the jth problem, the probability of correct solve is Pij .
At any time, the different between any two students’ programming time is not more than 1 hour. For example, if there are 3 students and there are 5 problems. The strategy {1,2,3,1,2}, {1,3,2,2,3} or {2,1,3,3,1} are all legal. But {1,1,3,2,3},{3,1,3,1,2} and
{1,2,3,1,1} are all illegal.
You should find a strategy to maximize the expected number of correctly solved problems, if you have know all probability
Input
The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.
The first line of each case contains two integers N ,M (1 ≤ N ≤ 10,1 ≤ M ≤ 1000),denoting the number of students and programming problem, respectively.
The next N lines, each lines contains M real numbers between 0 and 1 , the jth number in the ith line is Pij .
Output
For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then a single real number means the maximal expected number of correctly solved problems if this team follow the best strategy, to five digits
after the decimal point. Look at the output for sample input for details.
Sample Input
1
2 3
0.6 0.3 0.4
0.3 0.7 0.9
Sample Output
Source
2014 ACM/ICPC Asia Regional Shanghai Online
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可以转换成最小费用流问题，果断贴模板上去了。。。
//#pragma comment(linker, "/STACK:36777216")
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <climits>
#include <cassert>
#include <complex>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
#define typef int
#define typec double
const typef inff = 0x3f3f3f3f;
const typec infc = 0x3f3f3f3f;
const int E=1010;
const int N=100;
class network
{
public:
int nv, ne, pnt[E], nxt[E];
int vis[N], que[N], head[N], pv[N], pe[N];
typec cost, dis[E], d[N];typef flow, cap[E];
void addedge(int u, int v, typef c, typec w) {
pnt[ne] = v; cap[ne] = c;
dis[ne] = +w; nxt[ne] = head[u]; head[u] = ne++;
pnt[ne] = u; cap[ne] = 0;
dis[ne] = -w; nxt[ne] = head[v]; head[v] = ne++;
}
double mincost(int src, int sink) {
int i, k, f, r; typef mxf;
for (flow = 0, cost = 0; ; ) {
memset(pv, -1, sizeof(pv));
memset(vis, 0, sizeof(vis));
for (i = 0; i < nv; ++i) d[i] = infc;
d[src] = 0; pv[src] = src; vis[src] = 1;
for (f = 0, r = 1, que[0] = src; r != f; ) {
i = que[f++]; vis[i] = 0;
if (N == f) f = 0;
for (k = head[i]; k != -1; k = nxt[k])
if(cap[k] && dis[k]+d[i] < d[pnt[k]]){
d[pnt[k]] = dis[k] + d[i];
if (0 == vis[pnt[k]]) {
vis[pnt[k]] = 1;
que[r++] = pnt[k];
if (N == r) r = 0;
}
pv[pnt[k]]=i; pe[pnt[k]]=k;
}
}
if (-1 == pv[sink]) break;
for (k = sink, mxf = inff; k != src; k = pv[k])
if (cap[pe[k]] < mxf) mxf = cap[pe[k]];
flow += mxf; cost += d[sink] * mxf;
for (k = sink; k != src; k = pv[k]) {
cap[pe[k]] -= mxf; cap[pe[k] ^ 1] += mxf;
}
}
return cost;
}
void build(int v) {
nv = v; ne = 0;
memset(head, -1, sizeof(head));
}
};network g;
int m,n; //n 人数 m题数
double So[20][1010];
double solve(int x,int y){
// cout<<"ST SOLVE"<<endl;
double ret=0;
int Pro=y-x+1;
int nv=n+Pro+1;
// cout<<"Pro="<<Pro<<' '<<"nv="<<nv<<' '<<"n="<<n<<endl;
g.build(nv+1);
for(int i=1;i<=n;i++)
g.addedge(0,i,1,0);
for(int i=n+1;i<=n+Pro;i++)
g.addedge(i,nv,1,0);
for(int i=1;i<=n;i++)
for(int j=n+1;j<=n+Pro;j++){
g.addedge(i,j,1,-So[i][j-n-1+x]);
}
ret=g.mincost(0,nv);
return ret;
}
int main(){
#ifdef ONLINE_JUDGE
#else
freopen("in.txt","r",stdin);
#endif
int T,kase=0;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%lf",&So[i][j]);
}
}
double ans=0.0;int t=0;
for(int i=1;i<=m;i+=n){
ans+=solve(i,min(m,i+n-1));
}
printf("Case #%d: %.5lf
",++kase,-ans);
}
return 0;
}
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