HDU 5988 最小费用最大流
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题目
Coding Contest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3503 Accepted Submission(s): 825
Problem Description
A coding contest will be held in this university, in a huge playground. The whole playground would be divided into N blocks, and there would be M directed paths linking these blocks. The i-th path goes from the ui -th block to the vi -th block. Your task is to solve the lunch issue. According to the arrangement, there are si competitors in the i-th block. Limited to the size of table, bi bags of lunch including breads, sausages and milk would be put in the i-th block. As a result, some competitors need to move to another block to access lunch. However, the playground is temporary, as a result there would be so many wires on the path.
For the i-th path, the wires have been stabilized at first and the first competitor who walker through it would not break the wires. Since then, however, when a person go through the i - th path, there is a chance ofpi to touch
the wires and affect the whole networks. Moreover, to protect these wires, no more thanci competitors are allowed to walk through the i-th path.
Now you need to find a way for all competitors to get their lunch, and minimize the possibility of network crashing.
For the i-th path, the wires have been stabilized at first and the first competitor who walker through it would not break the wires. Since then, however, when a person go through the i - th path, there is a chance of
the wires and affect the whole networks. Moreover, to protect these wires, no more than
Now you need to find a way for all competitors to get their lunch, and minimize the possibility of network crashing.
Input
The first line of input contains an integer t which is the number of test cases. Then t test cases follow.
For each test case, the first line consists of two integers N (N ≤ 100) and M (M ≤ 5000). Each of the next N lines contains two integers si andbi (si , bi ≤ 200).
Each of the next M lines contains three integersui , vi and ci(ci ≤ 100) and a float-point number pi (0 < pi < 1).
It is guaranteed that there is at least one way to let every competitor has lunch.
For each test case, the first line consists of two integers N (N ≤ 100) and M (M ≤ 5000). Each of the next N lines contains two integers si and
Each of the next M lines contains three integers
It is guaranteed that there is at least one way to let every competitor has lunch.
Output
For each turn of each case, output the minimum possibility that the networks would break down. Round it to 2 digits.
Sample Input
14 42 00 33 00 31 2 5 0.53 2 5 0.51 4 5 0.53 4 5 0.5
Sample Output
0.50
解题思路
TTTTTTTTTTTTTTT哭了
思路概率要相乘所以转化成log,第一次不计算,所以额外加一条不消耗的边
关键是超时,超时,超时,用的kuangbin的板子
double判断要用1e-8来提高速度
还发现pow似乎比exp快一些
能注意的就这么多了,浪费我一晚上去改
如果还不行,抱歉,兄台换板子吧,你的最小费用最大流板子可能真的不行
贴上我靠运气卡过的代码
#include <cstdio>#include <iostream>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int MAXN = 10000;const int MAXM = 100000;const double eps = 1e-8;const double INF = 999999999.0;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++;}typedef pair<double , int> P; 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] = 0.0; vis[s] = true; q.push(s); while(!q.empty()) { int u = q.front(); q.pop();/// int u = tt.second; 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)> eps)) { 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int minCostMaxflow(int s,int t,double &cost){ int flow = 0; cost = 0.0; 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 * (double)Min; } flow += Min; } return flow;}int main(){ int T; int nn, m; scanf("%d", &T); while(T--) { scanf("%d%d",&nn ,&m); int A, B; init(nn + 1); for(int i = 1; i <= nn; i++) { scanf("%d%d", &A, &B); if(A == B) continue; if(A > B) { addedge(0, i, A - B, 0.0); continue; } if(A < B) { addedge(i, nn + 1, B - A, 0.0); continue ; } } int u, v, cap; double cost; for(int j = 0; j < m; j++) { scanf("%d%d%d%lf", &u, &v, &cap, &cost); double w = -log10(1.0 - cost); if(cap == 0) continue ; if(cap == 1) addedge(u, v, 1, 0.0); else if(cap - 1) { addedge(u, v, 1, 0.0); addedge(u, v, cap - 1, w); } } double ans = 0.0; minCostMaxflow(0, N, ans); ans = 1 - pow(10, -ans); printf("%.2f\n", ans); } return 0 ;}
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