Another Graph GameTime
limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 1691 Accepted Submission (s): 630
Problem Descriptionalice and Bob are playing a game on an undirected graph with n (n is even) nodes and M edges. Every node I have its own weight Wv, and every edge e have its own weight We.
They take turns to do the following operations. During each operation, either Alice or Bob can take one of the nodes from the graph that haven ' t been taken before. Alice goes first.
The scoring rule is:one person can get the bonus attached to a node if he/she has choosen that node before. One person can get the bonus attached to a edge if he/she has choosen both node that induced by the edge before.
You can assume Alice and Bob be intelligent enough and do operations optimally, both Alice and Bob's target is maximize t Heir Score-opponent ' s.
What's the final result for Alice-bob.
Inputmuilticases. The first line has a numbers n and M. (1 <= n <=, 0<=m<=105) The next line has n numbers from W1 to Wn Which Wi is the weight of node I. (| WI|<=109)
The next m lines, each of the line has three numbers u, V, W, (1≤u,v≤n,|w|<=109) The first 2 numbers is the and the nodes on the E Dge, and the last one are the weight on the edge.
Outputone line the final result.
Sample Input
4 09 8) 6 5
Sample Output
2
Source2013 multi-university Training Contest 5
#include <iostream> #include <cstdio> #include <string> #include <algorithm>using namespace std ; #define LL Long long#define maxn 100000 + 10int N, m;double a[maxn];int main () {while (~scanf ("%d%d", &n, &m)) { for (int i=1; i<=n; i++) scanf ("%LFD", &a[i]); int u, v; Double W; for (int i=1; i<=m; i++) { scanf ("%d%d%lf", &u, &v, &w); W/= 2; A[u] + = W; A[V] + = w; } Sort (a+1, a+1+n); Double sum1 = 0, sum2 = 0; for (int i=n; i>=1; i--) { if (I & 1) sum2 + = A[i]; else sum1 + = A[i]; } printf ("%.0lf\n", sum1-sum2); } return 0;}
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