P3275 [SCOI2011]糖果
快noip了我还在干什么啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊
来我们看这道题
根据条件建图， 因为求得是最小值， 所以要跑最长路qwq（这是我记住的QAQ
不想写了让我们直接看看题解吧！
P3275 [SCOI2011]糖果（five20的题解）
有环代表条件冲突， 输出“-1”
建源点的时候反向建会快（这就很妙了
1 #include<queue> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<iostream> 5 #define ri register int 6 #define ll long long 7 using namespace std; 8 const int sz = 200010; 9 int n, k, num = 0; 10 ll ans; 11 int head[sz], dis[sz], tot[sz]; 12 bool flag; 13 bool vis[sz]; 14 struct edge { 15 int nxt, to, dis; 16 }e[sz << 1]; 17 inline int read() { 18 char ch = getchar(); int x = 0, f = 1; 19 while(ch > '9' || ch < '0') {if(ch == '-') f = -1; ch = getchar();} 20 while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0' ; ch = getchar();} 21 return x * f; 22 } 23 void add(int from, int to, int dis) { 24 e[++num].nxt = head[from]; 25 e[num].to = to; 26 e[num].dis = dis; 27 head[from] = num; 28 } 29 void spfa() { 30 queue<int>q; 31 q.push(0); 32 vis[0] = 1; 33 while(!q.empty()) { 34 int u = q.front(); 35 q.pop(); 36 if(tot[u] == n - 1) { 37 printf("-1"); 38 exit(0); 39 } 40 vis[u] = 0; 41 tot[u]++; 42 for(ri i = head[u]; i; i = e[i].nxt) { 43 int v = e[i].to; 44 if(dis[v] < dis[u] + e[i].dis) { 45 dis[v] = dis[u] + e[i].dis; 46 if(!vis[v]) { 47 vis[v] = 1; 48 q.push(v); 49 } 50 } 51 } 52 } 53 } 54 int main() { 55 scanf("%d%d", &n, &k); 56 for(ri i = 1; i <= k; i++) { 57 int opt = read(), u = read(), v = read(); 58 if(opt == 1) { 59 add(v, u, 0); 60 add(u, v, 0); 61 } 62 else if(opt == 2) { 63 if(u == v) { 64 flag = 1; 65 break; 66 } 67 else add(u, v, 1); 68 } 69 else if(opt == 3) add(v, u, 0); 70 else if(opt == 4) { 71 if(u == v) { 72 flag = 1; 73 break; 74 } 75 else add(v, u, 1); 76 } 77 else if(opt == 5) add(u, v, 0); 78 } 79 if(flag == 1) { 80 printf("-1"); 81 return 0; 82 } 83 for(ri i = n; i >= 1; i--) add(0, i, 1); 84 spfa(); 85 for(ri i = 1; i <= n; i++) 86 ans += dis[i]; 87 printf("%lld", ans); 88 return 0; 89 } 90 /* 91 4 7 92 1 3 2 93 2 2 4 94 5 1 3 95 3 4 2 96 3 2 3 97 4 3 1 98 5 1 4 99 */