East China Jiaotong University 2015 ACM "double Base" program design contest 1003

Problem CTime limit:3000/1000ms (java/other) Memory limit:65535/32768k (Java/other) total submission (s): 225 Accep Ted Submission (s): 20font:times New Roman | Verdana | Georgiafont Size:←→problem Description "furnace Stone legend: Warcraft Heroes Biography" (Hearthstone:heroes of Warcraft, hereinafter referred to as the Furnace Stone legend) is a blizzard entertainment development of a set of card game ...
Simplified under the title: Two players A and B each have a number of cards (each card only attack and blood two properties, no additional attributes), each turn a card can only attack once,
and a card can only be attacked once, attack a card at the same time will be attacked by the attack card, the amount of blood to attack your card damage, when the blood volume is less than 1 o'clock, this card will be wiped out.
In the current round, a first attack, ask a attack, B on the left hand of the attack and the minimum amount of damage? Input the first line of T (T<=10), which represents the T-group data.
In each group of data, the first behavior n,m (0<n,m<8) represents N cards on field A, and M cards on the B field respectively.
In the next n rows, line I input aa[i],ab[i],aa[i] ab[i] respectively represents the attack and blood volume of the I card in field a.
In the next M line, line I input ba[i],bb[i],ba[i] bb[i] respectively represents the attack and blood volume of the I card on the B field.
(0<aa[i],ab[i],ba[i],bb[i]<=10) Output one row for each set of data, corresponding to the amount of damage left on the B field and the minimum value. Sample Input

11 32 18 27 26 2

Sample Output

13

Authormoonlike and real game is not the same, a each can only attack once and B each can only be attacked once. Then, the attackers in accordance with the low rank of attack to high order, the attack attacks high to low, the attack is equal, the blood is low to high order. And then we'll cycle.

#include <stdio.h>//#include <bits/stdc++.h> #include <string.h> #include <iostream> #include <math.h> #include <sstream> #include <set> #include <queue> #include <map> #include < vector> #include <algorithm> #include <limits.h> #define INF 0x3fffffff#define inf 0x3f3f3f3f#define Lson L,m,rt<<1#define Rson m+1,r,rt<<1|1#define LL long long#define ULL unsigned long longusing namespace Std;stru CT p{int x, y,} hehe[1000];struct c{int x, y,} he[1000];bool cmd1 (P a,p b) {return a.x<b.x;}    BOOL Cmd2 (C a,c b) {if (a.x==b.x) {return a.y<b.y;    } else {return a.x>b.x;    }}int Main () {int n,m;    int t;    int i,j;            while (cin>>t) {while (t--) {int ans=0;            cin>>n>>m;            for (i=0; i<n; i++) {cin>>hehe[i].x>>hehe[i].y;          } for (i=0; i<m; i++) {      cin>>he[i].x>>he[i].y;            } sort (HEHE,HEHE+N,CMD1);            Sort (HE,HE+M,CMD2); For (i=0, i<n; i++) {for (j=0; j<m; J + +) {if (HEHE[I].X&G                        t;=he[j].y&&he[j].x!=0) {he[j].x=0;                    Break            }}} for (i=0; i<m; i++) {ans+=he[i].x;        } cout<<ans<<endl; }} return 0;}

　　

East China Jiaotong University 2015 ACM "double Base" program design contest 1003