sed exceptio total icpc def i++ after ive pac

題目鏈接：https://vjudge.net/problem/HDU-3488

Tour

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 3720 Accepted Submission(s): 1777

Problem Description In the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M (M <= 30000) one-way roads connecting them. You are lucky enough to have a chance to have a tour in the kingdom. The route should be designed as: The route should contain one or more loops. (A loop is a route like: A->B->……->P->A.)
Every city should be just in one route.
A loop should have at least two cities. In one route, each city should be visited just once. (The only exception is that the first and the last city should be the same and this city is visited twice.)
The total distance the N roads you have chosen should be minimized.

Input An integer T in the first line indicates the number of the test cases.
In each test case, the first line contains two integers N and M, indicating the number of the cities and the one-way roads. Then M lines followed, each line has three integers U, V and W (0 < W <= 10000), indicating that there is a road from U to V, with the distance of W.
It is guaranteed that at least one valid arrangement of the tour is existed.
A blank line is followed after each test case.

Output For each test case, output a line with exactly one integer, which is the minimum total distance.

Sample Input 1 6 9 1 2 5 2 3 5 3 1 10 3 4 12 4 1 8 4 6 11 5 4 7 5 6 9 6 5 4

Sample Output 42

Source 2010 ACM-ICPC Multi-University Training Contest（6）——Host by BIT

Recommend zhouzeyong

題解：

代碼如下：

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 typedef long long LL;
  4 const int INF = 0x3f3f3f3f;
  5 const LL LNF = 9e18;
  6 const int mod = 1e9+7;
  7 const int MAXN = 2e2+10;
  8 
  9 int nx, ny;
 10 int g[MAXN][MAXN];
 11 int linker[MAXN], lx[MAXN], ly[MAXN];
 12 int slack[MAXN];
 13 bool visx[MAXN], visy[MAXN];
 14 
 15 bool DFS(int x)
 16 {
 17     visx[x] = true;
 18     for(int y = 1; y<=ny; y++)
 19     {
 20         if(visy[y]) continue;
 21         int tmp = lx[x] + ly[y] - g[x][y];
 22         if(tmp==0)
 23         {
 24             visy[y] = true;
 25             if(linker[y]==-1 || DFS(linker[y]))
 26             {
 27                 linker[y] = x;
 28                 return true;
 29             }
 30         }
 31         else
 32             slack[y] = min(slack[y], tmp);
 33     }
 34     return false;
 35 }
 36 
 37 int KM()
 38 {
 39     memset(linker, -1, sizeof(linker));
 40     memset(ly, 0, sizeof(ly));
 41     for(int i = 1; i<=nx; i++)
 42     {
 43         lx[i] = -INF;
 44         for(int j = 1; j<=ny; j++)
 45             lx[i] = max(lx[i], g[i][j]);
 46     }
 47 
 48     for(int x = 1; x<=nx; x++)
 49     {
 50         for(int i = 1; i<=ny; i++)
 51             slack[i] = INF;
 52         while(true)
 53         {
 54             memset(visx, 0, sizeof(visx));
 55             memset(visy, 0, sizeof(visy));
 56 
 57             if(DFS(x)) break;
 58             int d = INF;
 59             for(int i = 1; i<=ny; i++)
 60                 if(!visy[i])
 61                     d = min(d, slack[i]);
 62 
 63             for(int i = 1; i<=nx; i++)
 64                 if(visx[i])
 65                     lx[i] -= d;
 66             for(int i = 1; i<=ny; i++)
 67             {
 68                 if(visy[i]) ly[i] += d;
 69                 else slack[i] -= d;
 70             }
 71         }
 72     }
 73 
 74     int res = 0;
 75     for(int i = 1; i<=ny; i++)
 76         if(linker[i]!=-1)
 77             res += g[linker[i]][i];
 78     return res;
 79 }
 80 
 81 int main()
 82 {
 83     int T, n, m;
 84     scanf("%d", &T);
 85     while(T--)
 86     {
 87         scanf("%d%d", &n,&m);
 88         nx = ny = n;
 89         memset(g, 0, sizeof(g));
 90         for(int i = 1; i<=nx; i++)
 91             for(int j = 1; j<=ny; j++)
 92                 g[i][j] = -INF;
 93         for(int i = 1; i<=m; i++)
 94         {
 95             int u, v, w;
 96             scanf("%d%d%d", &u, &v, &w);
 97             g[u][v] = max(g[u][v], -w);
 98         }
 99 
100         printf("%d\n", -KM());
101     }
102 }

View Code

HDU3488 Tour —— 二分圖最大權匹配 KM算法