Problem Description
Wangpeng has N clothes, M pants and K shoes so theoretically he can have N×M×K different combinations of dressing.
One day he wears his pants Nike, shoes Adiwang to go to school happily. When he opens the door, his mom asks him to come back and switch the dressing. Mom thinks that pants-shoes pair is disharmonious because Adiwang is much better than Nike. After being asked to switch again and again Wangpeng figure out all the pairs mom thinks disharmonious. They can be only clothes-pants pairs or pants-shoes pairs.
Please calculate the number of different combinations of dressing under mom’s restriction.
One day he wears his pants Nike, shoes Adiwang to go to school happily. When he opens the door, his mom asks him to come back and switch the dressing. Mom thinks that pants-shoes pair is disharmonious because Adiwang is much better than Nike. After being asked to switch again and again Wangpeng figure out all the pairs mom thinks disharmonious. They can be only clothes-pants pairs or pants-shoes pairs.
Please calculate the number of different combinations of dressing under mom’s restriction.
Input
There are multiple test cases.
For each case, the first line contains 3 integers N,M,K(1≤N,M,K≤1000) indicating the number of clothes, pants and shoes.
Second line contains only one integer P(0≤P≤2000000) indicating the number of pairs which mom thinks disharmonious.
Next P lines each line will be one of the two forms“clothes x pants y” or “pants y shoes z”.
The first form indicates pair of x-th clothes and y-th pants is disharmonious(1≤x≤N,1 ≤y≤M), and second form indicates pair of y-th pants and z-th shoes is disharmonious(1≤y≤M,1≤z≤K).
Input ends with “0 0 0”.
It is guaranteed that all the pairs are different.
For each case, the first line contains 3 integers N,M,K(1≤N,M,K≤1000) indicating the number of clothes, pants and shoes.
Second line contains only one integer P(0≤P≤2000000) indicating the number of pairs which mom thinks disharmonious.
Next P lines each line will be one of the two forms“clothes x pants y” or “pants y shoes z”.
The first form indicates pair of x-th clothes and y-th pants is disharmonious(1≤x≤N,1 ≤y≤M), and second form indicates pair of y-th pants and z-th shoes is disharmonious(1≤y≤M,1≤z≤K).
Input ends with “0 0 0”.
It is guaranteed that all the pairs are different.
Output
For each case, output the answer in one line.
Sample Input
2 2 2
0
2 2 2
1
clothes 1 pants 1
2 2 2
2
clothes 1 pants 1
pants 1 shoes 1
0 0 0
Sample Output
8
6
5
Source
对于这题，看似跟容斥定理一样，不过P的值太大果断放弃。
后来想想只有三组结点，结点的个数为1000，不大果断使用使用map记录一个结点到另外一个结点的边。并且中间结点记录出度个数。
然后删边，最后使用加法原理直接加。
#include <stdio.h> #include <string.h> #define MAXN 1001 int N,M,K; int map[MAXN][MAXN]; int out[MAXN]; int main(int argc, char *argv[]) { char ch1[10],ch2[10]; int x,y; while( scanf("%d %d %d",&N,&M,&K)!=EOF ){ if(N==0 &&M==0 && K==0)break; memset(map,1,sizeof(map)); for(int i=1; i<=M; i++){ out[i]=K; } int P; scanf("%d",&P); while(P--){ scanf("%s %d %s %d",ch1,&x,ch2,&y); if(ch1[0]=='c' && ch2[0]=='p'){ map[x][y]=0; } if(ch1[0]=='p' && ch2[0]=='s'){ if(out[x]>0) out[x]--; } } __int64 sum=0; for(int i=1; i<=N; i++){ for(int j=1; j<=M; j++){ if(map[i][j]){ sum+=out[j]; } } } printf("%I64d ",sum); } return 0; }