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  • UVa 11374

    题目链接:https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=2369&mosmsg=Submission+received+with+ID+26582716

    可以修改一条边的权值,求起点到终点的最短路

    从起点和终点分别做一次最短路,枚举修改权值的边是哪一条,新的最短路长度就是两边的最短路加上新的权值

    还要输出路径,更新时记录前驱即可

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    typedef pair<int, int> pii;
    
    const int maxn = 10010; 
    const int INF = 0x3f3f3f3f;
    
    int n, s, t, m, k;
    int u[maxn], v[maxn], w[maxn];
    
    int h[maxn], cnt = 0;
    struct E{
    	int to, cost, next;
    }e[maxn];
    void add(int u, int v, int w){
    	e[++cnt].to = v;
    	e[cnt].cost = w;
    	e[cnt].next = h[u];
    	h[u] = cnt;
    }
    
    vector<int> path;
    int d[2][maxn];
    int pre[2][maxn];
    void dij(int S, int type){
    	priority_queue<pii, vector<pii>, greater<pii> > q;
    	d[type][S] = 0;
    	q.push(pii(0, S));
    	
    	while(!q.empty()){
    		pii p = q.top(); q.pop();
    		int u = p.second;
    		
    		if(p.first != d[type][u]) continue; 
    		
    		for(int i = h[u] ; i != -1 ; i = e[i].next){
    			int v = e[i].to;
    			if(d[type][v] > d[type][u] + e[i].cost){
    				d[type][v] = d[type][u] + e[i].cost;
    				pre[type][v] = u;
    				q.push(pii(d[type][v], v));
    			}
    		}
    	}
    }
    
    ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }
    
    int main(){
    	int flag = 0;
    	while(scanf("%d%d%d", &n, &s, &t) == 3){
    		if(flag) printf("
    ");
    		flag = 1;
    		path.clear();
    		memset(pre, 0, sizeof(pre));
    		memset(h, -1, sizeof(h)); cnt = 0;
    		
    		scanf("%d", &m);
    		for(int i = 1 ; i <= m ; ++i){
    			int u, v, w;
    			scanf("%d%d%d", &u, &v, &w);
    			add(u, v, w); add(v, u, w);
    		}
    		
    		scanf("%d", &k);
    		for(int i = 1 ; i <= k ; ++i){
    			scanf("%d%d%d", &u[i], &v[i], &w[i]);
    		} 
    		
    		memset(d, 0x3f, sizeof(d));
    		dij(s, 0);
    		dij(t, 1);
    		
    		int ans = d[0][t];
    		int sta = -1, stb = -1;
    		for(int i = 1 ; i <= k ; ++i){
    			if(d[0][u[i]] + d[1][v[i]] + w[i] < ans){
    				ans = d[0][u[i]] + d[1][v[i]] + w[i];
    				sta = u[i]; stb = v[i];
    			}
    			if(d[0][v[i]] + d[1][u[i]] + w[i] < ans){
    				ans = d[0][v[i]] + d[1][u[i]] + w[i];
    				sta = v[i]; stb = u[i];
    			}
    		}
    		
    		if(sta != -1){
    			int cur = sta;
    			while(1){
    				path.push_back(cur);
    				if(cur == s) break;
    				cur = pre[0][cur];
    			}
    			for(int i = path.size() - 1 ; i >= 0 ; --i){
    				printf("%d ", path[i]);
    			} 
    			path.clear();
    	
    			cur = stb;
    			while(1){
    				path.push_back(cur);
    				if(cur == t) break;
    				cur = pre[1][cur];
    			}
    			for(int i = 0 ; i <= path.size() - 1 ; ++i){
    				if(i > 0) printf(" ");
    				printf("%d", path[i]);
    			} printf("
    ");
    		} else{
    			int cur = t;
    			while(1){
    				path.push_back(cur);
    				if(cur == s) break;
    				cur = pre[0][cur];
    			}
    			
    			for(int i = path.size() - 1 ; i >= 0 ; --i){
    				if(i < path.size()-1) printf(" ");
    				printf("%d", path[i]);
    			} printf("
    ");
    		}
    		if(sta != -1) printf("%d
    ", sta);
    		else printf("Ticket Not Used
    ");
    		printf("%d
    ", ans);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/tuchen/p/15025570.html
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