标签:string == 匹配 str date() cstring queue tar return
https://vjudge.net/problem/UVA-11383
题意:
给定一个N×N矩阵,每个格子里都有一个正整数W(i,j)。你的任务是给每行确定一个整数row(i),每列也确定一个整数col(i),使得对于任意格子(i,j),w(i,j)<=row(i)+col(j)。所有的row(i)和col(i)只和应尽量小。
思路:
利用二分图最佳完美匹配当中的l(x)+l(y)>=w(i,j),直接用KM算法即可。
1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<algorithm> 5 #include<queue> 6 #include<cstdio> 7 using namespace std; 8 const int maxn=500+5; 9 10 int W[maxn][maxn], n; 11 int Lx[maxn]; 12 int Ly[maxn]; 13 int Left[maxn]; 14 bool S[maxn], T[maxn]; 15 16 bool Match(int i) 17 { 18 S[i] = true; 19 for (int j = 1; j <= n; j++) 20 { 21 if (Lx[i] + Ly[j] == W[i][j] && !T[j]) 22 { 23 T[j] = true; 24 if (!Left[j] || Match(Left[j])) 25 { 26 Left[j] = i; 27 return true; 28 } 29 } 30 } 31 return false; 32 } 33 34 void Update() 35 { 36 int a = 1 << 30; 37 for (int i = 1; i <= n; i++) 38 { 39 if (S[i]) 40 { 41 for (int j = 1; j <= n; j++) 42 { 43 if (!T[j]) 44 { 45 a = min(a, Lx[i] + Ly[j] - W[i][j]); 46 } 47 } 48 } 49 } 50 for (int i = 1; i <= n; i++) 51 { 52 if (S[i]) 53 Lx[i] -= a; 54 if (T[i]) 55 Ly[i] += a; 56 } 57 } 58 59 void KM() 60 { 61 for (int i = 1; i <= n; i++) 62 { 63 Left[i] = 0; 64 Lx[i] = 0; 65 Ly[i] = 0; 66 for (int j = 1; j <= n; j++) 67 { 68 Lx[i] = max(Lx[i], W[i][j]); 69 } 70 } 71 for (int i = 1; i <= n; i++) 72 { 73 while (true) 74 { 75 memset(S, 0, sizeof(S)); 76 memset(T, 0, sizeof(T)); 77 if (Match(i)) 78 break; 79 else 80 Update(); 81 } 82 } 83 } 84 85 int main() 86 { 87 //freopen("D:\\input.txt","r",stdin); 88 while(~scanf("%d",&n)) 89 { 90 for(int i=1;i<=n;i++) 91 { 92 for(int j=1;j<=n;j++) 93 scanf("%d",&W[i][j]); 94 } 95 int ans=0; 96 KM(); 97 for(int i=1;i<=n;i++) 98 { 99 printf("%d%c", Lx[i], i == n ? ‘\n‘ : ‘ ‘); 100 ans+=Lx[i]; 101 } 102 for(int i=1;i<=n;i++) 103 { 104 printf("%d%c", Ly[i], i == n ? ‘\n‘ : ‘ ‘); 105 ans+=Ly[i]; 106 } 107 printf("%d\n",ans); 108 } 109 return 0; 110 }
标签:string == 匹配 str date() cstring queue tar return
原文地址:http://www.cnblogs.com/zyb993963526/p/6883944.html