HDU 3488 2010多校联合6 图论-优快云博客

本文探讨了深度学习技术在音视频处理、AR特效、AI音视频处理等领域的应用，包括图像处理、音视频编解码、自然语言处理、区块链、隐私计算等关键技术。介绍了深度学习在音视频分割、语义识别、自动驾驶、AR增强现实等方面的实际案例，展示了其在现代科技发展中的重要作用。

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Tour

Time Limit: 3000/1000 MS (Java/Others)Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 725Accepted Submission(s): 363

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

刚开始看题表示完全木有思路。

后来看到别人用匹配。然后想了下，觉得有道理，就写了下，还好，1Y。

不过个人感觉还是没有领悟到精髓所在。。恩

先把代码贴出来，在仔细想想。恩

我的代码：

#include<stdio.h> #include<string.h> #define inf 99999999 int map[205][205]; int lx[205],ly[205]; bool x[205],y[205]; int link[205]; int n; bool dfs(int u) { int i; x[u]=true; for(i=1;i<=n;i++) { if(lx[u]+ly[i]==map[u][i]&&!y[i]) { y[i]=true; if(link[i]==-1||dfs(link[i])) { link[i]=u; return true; } } } return false; } int main() { int i,t,j,m; int a,b,l,k; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(i=1;i<=n;i++) for(j=1;j<=n;j++) map[i][j]=-inf; for(i=1;i<=m;i++) { scanf("%d%d%d",&a,&b,&l); if(map[a][b]<-l) map[a][b]=-l; } memset(x,0,sizeof(x)); memset(y,0,sizeof(y)); memset(link,-1,sizeof(link)); for(i=0;i<205;i++) lx[i]=inf; memset(ly,0,sizeof(ly)); for(k=1;k<=n;k++) { while(1) { memset(x,0,sizeof(x)); memset(y,0,sizeof(y)); if(dfs(k)) break; int d=inf; for(i=1;i<=n;i++) if(x[i]) for(j=1;j<=n;j++) if(!y[j]&&lx[i]+ly[j]-map[i][j]<d) d=lx[i]+ly[j]-map[i][j]; for(i=1;i<=n;i++) if(x[i]) lx[i]=lx[i]-d; for(i=1;i<=n;i++) if(y[i]) ly[i]=ly[i]+d; } } int ans=0; for(i=1;i<=n;i++) ans=ans+map[link[i]][i]; printf("%d\n",-ans); } return 0; }